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<cstdio>
const int N=5*1e5+10;
char s[N];
int a[N],n,m,sum[N],l,k,x,y,ans;
int ma(int a,int b){return a>b?a:b;}
int mi(int a,int b){return a>b?b:a;}
int main()
{
	scanf("%d%d%s",&n,&m,s);
	for(int i=1;i<=n;i++)a[i]=s[i-1]-'0';
	if(!a[1])l++;
	k=1;
	for(int i=2;i<=n;i++)
		if(a[i]==a[i-1])k++;
		else
		{
			a[++l]=k;
			k=1;
		}
	a[++l]=k;
	for(int i=1;i<=l;i++)
		sum[i]=sum[i-1]+a[i];
	for(int i=1;i<=l;i+=2)
		ans=ma(ans,sum[mi(l,i+2*m)]-sum[i-1]);
	printf("%d",ans);
	return 0;
}