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 main () {
  int n,k;
  cin>>n>>k;
  string a;
  cin>>a;
  vector<int> b(n);
  vector<int> c(n);
  int j=0;
  if (a[0]=='0') {
    j+=1;
  }
  
  for (int i=0;i<n-1;++i) {
    if (a[i]=='0' && a[i+1]=='1') {
      b[j]=i+1;
    }
    if (a[i]=='1' && a[i+1]=='0') {
      c[j]=i;
      j+=1;
    }
  }
  
  if (a[n-1]=='1') {
    c[j]=n-1;
  } else {
    b[j]=n-1;
    c[j]=n-1;
  }
  
  int ans=0;
  for (int l=0;l<=j;++l) {
    for (int t=0;t<=k;++t) {
    	int p=c[l+t]-b[l]+1;
    	ans=max(ans,p);
    }
  }
  cout<<ans<<endl;
}