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 <iostream>
#include <algorithm>
#include <string>
#include <vector>
using namespace std;
int main(void){
  int n,k,i,bi,hs=0,max;
  string s;
  vector<int> w;
  cin >> n >> k >> s;
  if (s[0]=='0') w.push_back(0);
  i=0;
  while (i<n){
    bi=i;
    while (i<n && s[i+1]==s[i]) i++;
    w.push_back(i-bi+1);
    i++;
  }
  if (s[n-1]=='0') w.push_back(0);
  if (w.size()<=k*2+1){
    cout << n << endl;
    return 0;
  }
  for (i=0;i<k*2+1;i++) hs+=w[i];
  max=hs;
  for (i=0;i+k*2+2<w.size();i+=2){
    hs+=w[i+k*2+1]+w[i+k*2+2]-w[i]-w[i+1];
    if (max<hs) max=hs;
  }
  cout << max << endl;
  return 0;
}