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;

#define rep(i,m,n) for(int i=m;i<n;i++)


typedef long long ll;


int main(){
   cin.tie(nullptr);
  ios::sync_with_stdio(false);

 ll n,k;
  cin >> n >> k;
  string s;
  cin >> s;
  vector<ll> index;
  index.push_back(0);
  
  rep(i,1,n) if(s[i]!=s[i-1]) index.push_back(i);
  
    rep(i,0,2*k+2) index.push_back(n);
  
  ll res=0;
  rep(i,0,index.size()){
   if(index[i]==n) break;
    res=max(res,index[i+2*k+ll(s[index[i]]-'0')]-index[i]);
  }
  cout << res << endl;
  
}
