Codeforces Round #624 (Div. 3) C - Perform the Combo

Perform the Combo

Perform the Combo

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You want to perform the combo on your opponent in one popular fighting game. The combo is the string ss consisting of nn lowercase Latin letters. To perform the combo, you have to press all buttons in the order they appear in ss. I.e. if s=s="abca" then you have to press 'a', then 'b', 'c' and 'a' again.

You know that you will spend mm wrong tries to perform the combo and during the ii-th try you will make a mistake right after pipi-th button (1≤pi<n1≤pi<n) (i.e. you will press first pipi buttons right and start performing the combo from the beginning). It is guaranteed that during the m+1m+1-th try you press all buttons right and finally perform the combo.

I.e. if s=s="abca", m=2m=2 and p=[1,3]p=[1,3] then the sequence of pressed buttons will be 'a' (here you're making a mistake and start performing the combo from the beginning), 'a', 'b', 'c', (here you're making a mistake and start performing the combo from the beginning), 'a' (note that at this point you will not perform the combo because of the mistake), 'b', 'c', 'a'.

Your task is to calculate for each button (letter) the number of times you'll press it.

You have to answer tt independent test cases.

Input

The first line of the input contains one integer tt (1≤t≤1041≤t≤104) — the number of test cases.

Then tt test cases follow.

The first line of each test case contains two integers nn and mm (2≤n≤2⋅1052≤n≤2⋅105, 1≤m≤2⋅1051≤m≤2⋅105) — the length of ss and the number of tries correspondingly.

The second line of each test case contains the string ss consisting of nn lowercase Latin letters.

The third line of each test case contains mm integers p1,p2,…,pmp1,p2,…,pm (1≤pi<n1≤pi<n) — the number of characters pressed right during the ii-th try.

It is guaranteed that the sum of nn and the sum of mm both does not exceed 2⋅1052⋅105 (∑n≤2⋅105∑n≤2⋅105, ∑m≤2⋅105∑m≤2⋅105).

It is guaranteed that the answer for each letter does not exceed 2⋅1092⋅109.

Output

For each test case, print the answer — 2626 integers: the number of times you press the button 'a', the number of times you press the button 'b', ……, the number of times you press the button 'z'.

Example

input

3
4 2
abca
1 3
10 5
codeforces
2 8 3 2 9
26 10
qwertyuioplkjhgfdsazxcvbnm
20 10 1 2 3 5 10 5 9 4

output

4 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 9 4 5 3 0 0 0 0 0 0 0 0 9 0 0 3 1 0 0 0 0 0 0 0 
2 1 1 2 9 2 2 2 5 2 2 2 1 1 5 4 11 8 2 7 5 1 10 1 5 2 

Note

The first test case is described in the problem statement. Wrong tries are "a", "abc" and the final try is "abca". The number of times you press 'a' is 44, 'b' is 22 and 'c' is 22.

In the second test case, there are five wrong tries: "co", "codeforc", "cod", "co", "codeforce" and the final try is "codeforces". The number of times you press 'c' is 99, 'd' is 44, 'e' is 55, 'f' is 33, 'o' is 99, 'r' is 33 and 's' is 11.

水题，我觉得没必要讲细节了，详见代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rg register ll
#define inf 2147483647
#define lb(x) (x&(-x))
ll sz[200005],n;
template <typename T> inline void read(T& x)
{
    x=0;char ch=getchar();ll f=1;
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}x*=f;
}
inline ll query(ll x){ll res=0;while(x){res+=sz[x];x-=lb(x);}return res;}
inline void add(ll x,ll val){while(x<=n){sz[x]+=val;x+=lb(x);}}//第x个加上val
ll t,cnt[30],sum[200005][30],ans[30];
int main()
{
	cin>>t;
    for(rg i=1;i<=t;i++)
    {
        ll n,m;
        string s;
        cin>>n>>m;
        cin>>s;
        for(rg k=0;k<26;k++)sum[0][k]=0;
        memset(ans,0,sizeof(ans));
        for(rg j=0;s[j];j++)
        {
            for(rg k=0;k<26;k++)sum[j+1][k]=sum[j][k];
            sum[j+1][s[j]-'a']++;
            //cout<<j+1<<" "<<s[j]-'a'<<" "<<sum[j+1][s[j]-'a']<<endl;
        }
        for(rg i=0;i<m;i++)
        {
            ll x;
            cin>>x;
            for(rg j=0;j<26;j++)
            {
                if(sum[x][j])ans[j]+=sum[x][j];
            }
        }
        for(rg j=0;j<26;j++)
        {
            if(sum[n][j])ans[j]+=sum[n][j];
        }
        for(rg i=0;i<26;i++)
        {
            i==26?cout<<ans[i]<<endl:cout<<ans[i]<<" ";
        }
    }
   // while(1)getchar();
    return 0;
    
}