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<pair<int,int>> T(1,{1,1});
  bool z=true;
  int s;
  for(int i=0;i<N;i++){
    if(z&&S[i]=='1'){
      s=i; z=false;
    }
    if(!z&&S[i]=='0'){
      T.push_back({s+1,i});
      z=true;
    }
  }
  if(!z)
    T.push_back({s+1,N});
  T.push_back({N,N});
  int ts=T.size();
  int ans=0,ansc;
  for(int i=0;i<ts;i++){
    //cout<<T[i].first<<" "<<T[i].second<<endl;
    if(i+K<ts){
      ansc=T[i+K].second-T[i].first+1;
      ans=max(ans,ansc);
    }
  }
  if(ans)
    cout<<ans<<endl;
  else
    cout<<N<<endl;
}