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(){
  #define int long
  int n,k,pos;
  string s;
  cin>>n>>k>>s;
  vector<int> v(1,0);
  if(s[0]=='1'){v[0]++;pos=0;}
  else {v.emplace_back(1);pos=1;}
  for(int i=1;i<n;++i){
    if(s[i]==s[i-1])v[pos]++;
    else {
      v.emplace_back(1);
      pos++;
    }
  }
  if(s[n-1]=='0')v.emplace_back(0);
  n=v.size();
  for(int i=1;i<n;++i)v[i]+=v[i-1];
  k=min(k,(int)v.size()/2);
  int res=v[2*k];
  for(int i=2*k+2;i<(int)v.size();i+=2){
    res=max(res,v[i]-v[i-2*k-1]);
  }
  cout<<res<<endl;
  return 0;
}