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() {
  long n,k,ans;
  string s;
  vector<long> a;
  cin >> n >> k >> s;
  
  if(s[0]=='0') a.push_back(0);
  a.push_back(1);
  for(long i=1;i<n;i++) {
    if(s[i-1]==s[i]) a[a.size()-1]++;
    else a.push_back(1);
  }
  if(s[n-1]=='0') a.push_back(0);

  if(a.size()<=k*2+1) {
    ans=n;
  } else {
    long x=0;
    for(long i=0;i<k*2+1;i++) x+=a[i];
    ans=x;
    for(long i=0;i<(a.size()-k*2+1)/2;i++) {
      x-=a[i*2]+a[i*2+1];
      x+=a[k*2+i*2+1]+a[k*2+i*2+2];
      ans=max(ans,x);
    }
  }
  cout << ans << endl;
}