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 s;cin >> s;
  vector<int> v;
  v.push_back(0);
  for(int i=0;i<n-1;i++){
    if(s[i]!=s[i+1]){
      v.push_back(i+1);
    }
  }
  v.push_back(n);
  int ans=0;
  int u=v.size();
  for(int i=0;i<v.size();i++){
    if(s[v[i]]=='1'){
      ans=max(ans,v[min(i+2*k+1,u-1)]-v[i]);
    }
    else{
      ans=max(ans,v[min(i+2*k,u-1)]-v[i]);
    }
  }
  cout << ans << endl;
}