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,L[N],R[N],l,x,y,ans;
int ma(int a,int b){return a>b?a:b;}
int main()
{
	scanf("%d%d%s",&n,&m,s);
	for(int i=1;i<=n;i++)a[i]=s[i-1]-'0';
	for(int i=1;i<=n;i++)
		if(a[i]==0)
		{
			L[++l]=i;
			for(;i<=n;i++)if(a[i])break;
			i--;
			R[l]=i;
		}
	for(int i=1;i+m-1<=l;i++)
	{
		x=L[i]-1;
		y=R[i+m-1]+1;
		while(a[x])x--;
		while(a[y])y++;
		ans=ma(ans,y-x-1);
	}
	if(m>=l)ans=n;
	printf("%d",ans);
	return 0;
}