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 <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <utility>
#define maxn 100005
using namespace std;
typedef long long ll;
int n,k;
char s[maxn];
int main(){
  scanf("%d%d",&n,&k);
  scanf("%s",s + 1);
  s[0] = '1';
  int cnt = 0,j = 0,ans = 0;
  for(int i = 1;i <= n;i++){
    while(j <= n && cnt <= k){
      if(s[j] == '1' && s[j + 1] == '0') cnt++;
      j++;
    }
    j--;
    if(s[j] == '1' && s[j + 1] == '0') cnt--;
    ans = max(ans,j - i + 1);
    if(s[i] == '0' && s[i + 1] == '1') cnt--;
  }
  printf("%d\n",ans);
  return 0;
}
