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>
#include <vector>
#include <algorithm>
using namespace std;

int n,k,a[100005];
vector <int> v,av;

int main(void){
	scanf("%d %d",&n,&k);
	for(int i=1;i<=n;i++)
	scanf("%1d",&a[i]);
	if(a[1]==0)v.push_back(0);
	int len=0;
	for(int i=1;i<=n;i++){
		if(i>1&&a[i]!=a[i-1]){
			v.push_back(len);
			len=1;
		}
		else
		len++;
	}
	v.push_back(len);
	av.resize(v.size()+1);
	av[0]=0;
	for(int i=0;i<v.size();i++)
	av[i+1]=av[i]+v[i];
	int ans=0;
	for(int i=0;i<av.size();i+=2){
		int ss=av[min((int)v.size(),i+2*k+1)]-av[i];
		ans=max(ans,ss);
	}
	printf("%d\n",ans);
}