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>
#define pb push_back
using namespace std;

const int N=1e5+10;
int n,k;
char buf[N];
string str;
int cnt,pre[N];
vector<int>v;
int maxn=1;

int main(){
	scanf("%d%d\n%s",&n,&k,buf);
	str=(string)buf;
	cnt=0;
	if(str[0]=='0') v.pb(0);
	for(int i=0;i<str.size();i++)
		if(i>0&&str[i]!=str[i-1]) {
			v.pb(cnt);
			cnt=1;
		} else cnt++;
	v.pb(cnt);
	for(int i=0;i<v.size();i++)
		pre[i+1]=pre[i]+v[i];
	for(int i=0;i<v.size();i+=2){
		cnt=pre[min(int(v.size()),int(i+k*2+1))]-pre[i];
		maxn=max(maxn,cnt);
	}
	printf("%d\n",maxn);
	return 0;
}