Codeforces Round #625 (Div. 2, based on Technocup 2020 Final Round) C. Remove Adjacent

C. Remove Adjacent

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a string ss consisting of lowercase Latin letters. Let the length of ss be |s||s|. You may perform several operations on this string.

In one operation, you can choose some index ii and remove the ii-th character of ss (sisi) if at least one of its adjacent characters is the previous letter in the Latin alphabet for sisi. For example, the previous letter for b is a, the previous letter for s is r, the letter a has no previous letters. Note that after each removal the length of the string decreases by one. So, the index ii should satisfy the condition 1≤i≤|s|1≤i≤|s| during each operation.

For the character sisi adjacent characters are si−1si−1 and si+1si+1. The first and the last characters of ss both have only one adjacent character (unless |s|=1|s|=1).

Consider the following example. Let s=s= bacabcab.

1. During the first move, you can remove the first character s1=s1= b because s2=s2= a. Then the string becomes s=s= acabcab.
2. During the second move, you can remove the fifth character s5=s5= c because s4=s4= b. Then the string becomes s=s= acabab.
3. During the third move, you can remove the sixth character s6=s6='b' because s5=s5= a. Then the string becomes s=s= acaba.
4. During the fourth move, the only character you can remove is s4=s4= b, because s3=s3= a (or s5=s5= a). The string becomes s=s= acaa and you cannot do anything with it.

Your task is to find the maximum possible number of characters you can remove if you choose the sequence of operations optimally.

Input

The first line of the input contains one integer |s||s| (1≤|s|≤1001≤|s|≤100) — the length of ss.

The second line of the input contains one string ss consisting of |s||s| lowercase Latin letters.

Output

Print one integer — the maximum possible number of characters you can remove if you choose the sequence of moves optimally.

Examples

input

Copy

8
bacabcab

output

Copy

4

input

Copy

4
bcda

output

Copy

3

input

Copy

6
abbbbb

output

Copy

5

Note

The first example is described in the problem statement. Note that the sequence of moves provided in the statement is not the only, but it can be shown that the maximum possible answer to this test is 44.

In the second example, you can remove all but one character of ss. The only possible answer follows.

1. During the first move, remove the third character s3=s3= d, ss becomes bca.
2. During the second move, remove the second character s2=s2= c, ss becomes ba.
3. And during the third move, remove the first character s1=s1= b, ss becomes a.

题意：给定一个字符串，若字符串中的某个字符的前一个或者后一个是其字典序的前一个字母就可以去掉这个字符，问最多能去掉多少个字符

思路：贪心，每次去掉可以去掉字符的最大字典序的字母，可以脑补一下，如果每次都不去掉当前可以去掉的最大字典序的字母，有可能会导致策略不优，很明显嘛，比如4 bcda，你先去掉c，那么就是bda，只有1，而答案是3，这个需要多想想，其实也很显然的

#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
string s,ans[105];
int main()
{
	cin>>n>>s;
	ll tot=1,cnt=0;
	ans[tot]=s;
	while(1)
	{	
		vector<char>v;
		for(rg i=0;ans[tot][i];i++)
		{
			if(i==0)
			{
				if(ans[tot][i]-ans[tot][i+1]==1)v.push_back(ans[tot][i]);
				continue;
			}
			if(i==ans[tot].length()-1)
			{
				if(ans[tot][i]-ans[tot][i-1]==1)v.push_back(ans[tot][i]);
				continue;
			}
			if((ans[tot][i]-ans[tot][i-1]==1)||(ans[tot][i]-ans[tot][i+1]==1))v.push_back(ans[tot][i]);
			
		}
		//cout<<ans[tot]<<endl;
		if(!v.size())break;
		else
		{	
			sort(v.begin(),v.end());
			char ansc=v[v.size()-1];
			ll flag=0;
			for(rg i=0;ans[tot][i];i++)
			{
				if(ans[tot][i]==ansc&&!flag&&((ans[tot][i]-ans[tot][i-1]==1)||(ans[tot][i]-ans[tot][i+1]==1)))
				{
					flag=1;
					continue;
				}
				else  ans[tot+1]+=ans[tot][i];
			}
		}
		tot++;
		//cout<<ans[tot]<<endl;
		
		cnt++;
	}
	cout<<cnt<<endl;
   // while(1)getchar();
    return 0;
    
}