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;
  string s;
  cin >> n >> k >> s;
  
  vector<int> v;
  char prev = '1';
  int curr = 0;
  for (int i = 0; i <= s.size(); ++i) {
    if (s[i] != prev) {
      v.push_back(curr);
      prev = s[i];
      curr = 0;
    }
    ++curr;
  }
  
  vector<int> p(v.size() + 1);
  partial_sum(v.begin(), v.end(), p.begin() + 1);
  
  int mx = 0;
  for (int l = 0; l < v.size(); l += 2) {
    int r = min<int>(l + 2 * k + 1, v.size());
    mx = max(mx, p[r] - p[l]);
  }
  cout << mx << endl;
}