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;
int d[200000],a[200000],sum[200000];
int main()
{
	int n,K;
	cin>>n>>K;
	for(int i=0;i<n;i++)
	{
		scanf("%1d",&d[i+1]);
	}
	for(int i=1;i<=n;i++)
	a[i]=(d[i]^d[i-1]);
	sum[0]=0;
	for(int i=1;i<=n;i++)
	{
		sum[i]=sum[i-1]+a[i];
	}
	int b=0,Max=0;
	for(int i=1;i<=n;i++)
	{
		b^=a[i];
		int x=upper_bound(sum,sum+n+1,K*2-(!b)+sum[i])-sum;
		Max=max(Max,x-i);
	}
	cout<<Max<<"\n";
	return 0;
}