N people are arranged in a row from left to right.

You are given a string S of length N consisting of `0` and `1`, and a positive integer K.

The i-th person from the left is standing on feet if the i-th character of S is `0`, and standing on hands if that character is `1`.

You will give the following direction at most K times (possibly zero):

Direction: Choose integers l and r satisfying 1 \leq l \leq r \leq N, and flip the l-th, (l+1)-th, ..., and r-th persons. That is, for each i = l, l+1, ..., r, the i-th person from the left now stands on hands if he/she was standing on feet, and stands on feet if he/she was standing on hands.

Find the maximum possible number of consecutive people standing on hands after at most K directions.

Constraints

* N is an integer satisfying 1 \leq N \leq 10^5.
* K is an integer satisfying 1 \leq K \leq 10^5.
* The length of the string S is N.
* Each character of the string S is `0` or `1`.

Input

Input is given from Standard Input in the following format:


N K
S


Output

Print the maximum possible number of consecutive people standing on hands after at most K directions.

Examples

Input

5 1
00010


Output

4


Input

14 2
11101010110011


Output

8


Input

1 1
1


Output

1
#include <bits/stdc++.h>
using namespace std;
#define mx 100001
typedef long long L;
L n,k,mDis=0,bck=-1,dist=0;
char str[mx+5];
queue<L>q;
void solve(){
	L i,flg=0;
	for(i=0;i<n;i++){
		if(str[i]=='0'){
			while(i<n && str[i]=='0')i++;
			q.push(i-1);
			while(i<n&&str[i]=='1')i++;
			i--;
			if(q.size()==k){flg=1;
				mDis=(i-bck);
				dist=max(mDis,dist);
				bck=q.front();q.pop();
			}
		}
	}
	if(flg==0){
		dist=n;
	}
}

int main()
{
	scanf("%lld%lld",&n,&k);
	scanf("%s",&str);
	solve();
	printf("%lld\n",dist);
	return 0;
}
