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;

const int N=1e6+5;
int n,k,tl;
pair<int,int> lis[N];
char s[N];

void init()
{
	scanf("%d%d",&n,&k);int i;
	scanf("%s",s+1);lis[++tl]=make_pair(0,0);
	for(i=1;i<=n;++i)
	{
		if(s[i]=='1') continue;
		int r=i;
		while(s[r]==s[i]&&r<=n) ++r;--r;
		lis[++tl]=make_pair(i,r);
		i=r;
	}
	lis[++tl]=make_pair(n+1,n+1);
}

void work()
{
	if(tl-2<=k)
	{
		printf("%d\n",n);
		exit(0);
	}
	int i,ans=0;
	for(i=2;i<=tl-k;++i) ans=max(ans,lis[i+k].first-lis[i-1].second-1);
	printf("%d\n",ans);
}

int main()
{
	init();work();
	return 0;
}