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 <iostream>
#include<bits/stdc++.h>
#define ll long long
#define   ull unsigned long long 
const int MAX=1e6+9;
const int mod=1e9+7;
using namespace std;
int cnt[MAX];

int main(){
	int n ,k;
	cin>>n>>k;
	string s;
	cin>>s;
	int l=0 , cnt=0;
	int mx=1;
	for(int i=0;i<s.size();i++){
		if(s[i]==s[i-1]&&s[i]=='0')
			continue;
		if(s[i]=='0')
			cnt++;
		while(cnt>k&&l<i){
			if(s[l]=='0'){
					while(s[l]=='0')
						l++;
				cnt--;
			}
			else
				l++;
		}
		//cout<<i<<"\n";
		if(mx<(i-l+1))
			mx=i-l+1;
	}
	if(mx<(n-1-l+1))
		mx=n-1-l+1;
	cout<<mx;
	
}