# A. Mike and palindrome

time limit per test:2 seconds

memory limit per test:256 megabytes

input:standard input

output:standard output

Mike has a string s consisting of only lowercase English letters. He wants to change exactly one character from the string so that the resulting one is a palindrome.

A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not.

Input

The first and single line contains string s (1 ≤ |s| ≤ 15).

Output

Print "YES" (without quotes) if Mike can change exactly one character so that the resulting string is palindrome or "NO" (without quotes) otherwise.

Examples

Input

`abccaa`

Output

`YES`

Input

`abbcca`

Output

`NO`

Input

`abcda`

Output

```YES

``` 1 #include <bits/stdc++.h>
2 using namespace std;
3 int main()
4 {
5     char s;
6     while(cin>>s)
7     {
8         int len=strlen(s);
9         for(int i=0;i<len;i++)
10         {
11             for(int j='a';j<='z';j++)
12             {
13                 if(s[i]==j)
14                     continue;
15             }
16         }
17         int ans=0;
18         if(len%2!=0)
19         {
20             for(int i=0;i<(len)/2;i++)
21             {
22                 if(s[i]==s[len-i-1])
23                     ans++;
24             }
25             if(ans==(len)/2-1||ans==len/2)
26                 cout<<"YES"<<endl;
27             else cout<<"NO"<<endl;
28
29         }
30         else if(len%2==0)
31         {
32         for(int i=0;i<(len)/2;i++)
33         {
34                 if(s[i]==s[len-i-1])
35                     ans++;
36         }
37             if(ans==(len)/2-1)
38                 cout<<"YES"<<endl;
39             else cout<<"NO"<<endl;
40         }
41     }
42     return 0;
43 }```

# B. Mike and strings

time limit per test：2 seconds

memory limit per test：256 megabytes

input：standard input

output：standard output

Mike has n strings s1, s2, ..., sn each consisting of lowercase English letters. In one move he can choose a string si, erase the first character and append it to the end of the string. For example, if he has the string "coolmike", in one move he can transform it into the string "oolmikec".

Now Mike asks himself: what is minimal number of moves that he needs to do in order to make all the strings equal?

Input

The first line contains integer n (1 ≤ n ≤ 50) — the number of strings.

This is followed by n lines which contain a string each. The i-th line corresponding to string si. Lengths of strings are equal. Lengths of each string is positive and don't exceed 50.

Output

Print the minimal number of moves Mike needs in order to make all the strings equal or print  - 1 if there is no solution.

Examples

Input

```4
xzzwo
zwoxz
zzwox
xzzwo```

Output

`5`

Input

```2
molzv
lzvmo```

Output

`2`

Input

```3
kc
kc
kc```

Output

`0`

Input

```3
aa
aa
ab```

Output

`-1`

Note

In the first sample testcase the optimal scenario is to perform operations in such a way as to transform all strings into "zwoxz".

``` 1 #include <bits/stdc++.h>
2 using namespace std;
3 void change(string &s)
4 {
5     s=s.substr(1)+s.substr(0,1);//字符串截取函数,这里的目的就是一个一个换字符，找最小次数
6 }
7 int main()
8 {
9     int n;
10     while(cin>>n)
11     {
12         string s[n];
13         for(int i=0;i<n;i++)
14             cin>>s[i];
15         int ans=1e+8;
16         int sum;
17         for(int i=0;i<n;i++)
18         {
19             sum=0;
20             for(int j=0;j<n;j++)
21             {
22                 string t=s[j];
23                 while(t!=s[i])
24                 {
25                     change(t);
26                     sum++;
27                     if(sum>3000)
28                     {
29                         ans=-1;
30                         break;
31                     }
32                 }
33             }
34             ans=min(ans,sum);
35         }
36         cout<<ans<<endl;
37     }
38     return 0;
39 }```

# C. Mike and gcd problem

time limit per test：2 seconds

memory limit per test：256 megabytes

input：standard input

output：standard output

Mike has a sequence A = [a1, a2, ..., an] of length n. He considers the sequence B = [b1, b2, ..., bn] beautiful if the gcd of all its elements is bigger than 1, i.e.

.

Mike wants to change his sequence in order to make it beautiful. In one move he can choose an index i (1 ≤ i < n), delete numbers ai, ai + 1 and put numbers ai - ai + 1, ai + ai + 1 in their place instead, in this order. He wants perform as few operations as possible. Find the minimal number of operations to make sequence A beautiful if it's possible, or tell him that it is impossible to do so.

is the biggest non-negative number d such that d divides bi for every i (1 ≤ i ≤ n).

Input

The first line contains a single integer n (2 ≤ n ≤ 100 000) — length of sequence A.

The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — elements of sequence A.

Output

Output on the first line "YES" (without quotes) if it is possible to make sequence A beautiful by performing operations described above, and "NO" (without quotes) otherwise.

If the answer was "YES", output the minimal number of moves needed to make sequence A beautiful.

Examples

Input

```2
1 1```

Output

```YES
1```

Input

```3
6 2 4```

Output

```YES
0```

Input

```2
1 3```

Output

```YES
1```

Note

In the first example you can simply make one move to obtain sequence [0, 2] with

.

In the second example the gcd of the sequence is already greater than 1.

``` 1 #include <bits/stdc++.h>
2 using namespace std;
3 const int maxn=1e+7;
4 int gcd(int a,int b)
5 {
6     return b==0?a:gcd(b,a%b);
7 }
8 int a[maxn];
9 int main()
10 {
11     int n,i;
12     cin>>n;
13     for(int i=0;i<n;i++)
14     cin>>a[i];
15     int gccd=a;
16     for(i=1;i<n;i++)
17     {
18         gccd=gcd(gccd,a[i]);
19         if(gccd==1)
20             break;
21     }
22     if(i==n)
23     {
24         cout<<"YES"<<endl<<0<<endl;
25         return 0;
26     }
27     else
28     {
29         int c=0,cnt=0;
30         a[n]=0;
31         for(i=0;i<=n;i++)
32         {
33             if(a[i]%2!=0)
34                 cnt++;//求这组数中有多少个奇数
35             else
36             {
37                 c+=(cnt/2);
38                 cnt%=2;
39                 if(cnt!=0)
40                     c+=2;//一个奇数要变2次到偶数
41                 cnt=0;
42             }
43         }
44         cout<<"YES"<<endl<<c<<endl;
45     }
46     return 0;
47 }```

# D. Mike and distribution

time limit per test：2 seconds

memory limit per test：256 megabytes

input：standard input

output：standard output

Mike has always been thinking about the harshness of social inequality. He's so obsessed with it that sometimes it even affects him while solving problems. At the moment, Mike has two sequences of positive integers A = [a1, a2, ..., an] and B = [b1, b2, ..., bn] of length n each which he uses to ask people some quite peculiar questions.

To test you on how good are you at spotting inequality in life, he wants you to find an "unfair" subset of the original sequence. To be more precise, he wants you to select k numbers P = [p1, p2, ..., pk] such that 1 ≤ pi ≤ n for 1 ≤ i ≤ k and elements in P are distinct. Sequence P will represent indices of elements that you'll select from both sequences. He calls such a subset P "unfair" if and only if the following conditions are satisfied: 2·(ap1 + ... + apk) is greater than the sum of all elements from sequence A, and 2·(bp1 + ... + bpk) is greater than the sum of all elements from the sequence B. Also, k should be smaller or equal to

because it will be to easy to find sequence P if he allowed you to select too many elements!

Mike guarantees you that a solution will always exist given the conditions described above, so please help him satisfy his curiosity!

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of elements in the sequences.

On the second line there are n space-separated integers a1, ..., an (1 ≤ ai ≤ 109) — elements of sequence A.

On the third line there are also n space-separated integers b1, ..., bn (1 ≤ bi ≤ 109) — elements of sequence B.

Output

On the first line output an integer k which represents the size of the found subset. k should be less or equal to

.

On the next line print k integers p1, p2, ..., pk (1 ≤ pi ≤ n) — the elements of sequence P. You can print the numbers in any order you want. Elements in sequence P should be distinct.

Example

Input

`5 8 7 4 8 3 4 2 5 3 7`

Output

`3 1 4 5`

``` 1 #include <bits/stdc++.h>
2 using namespace std;
3 const int maxn=1e+6;
4 struct node
5 {
6     int a,b,j;
7 }p[maxn];
8 bool cmp(node c,node d)
9 {
10     if(c.a!=d.a)
11         return c.a>d.a;
12     else return c.b>d.b;
13 }
14 int main()
15 {
16     int n;
17     while(scanf("%d",&n)!=EOF)
18     {
19         for(int i=1;i<=n;i++)
20         {
21             scanf("%d",&p[i].a);
22             p[i].j=i;
23         }
24         for(int i=1;i<=n;i++)
25             scanf("%d",&p[i].b);;
26         sort(p+1,p+1+n,cmp);
27         printf("%d\n",n/2+1);
28         printf("%d ",p.j);
29         for(int i=1;i<=(n-1)/2;i++)
30         {
31             if(p[2*i].b>p[2*i+1].b)
32                 printf("%d ",p[2*i].j);
33             else printf("%d ",p[2*i+1].j);
34         }
35         if(n%2==0)
36             printf("%d ",p[n].j);
37         printf("\n");
38     }
39     return 0;
40 }```

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