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() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n, k, m = 0;
  string s;
  cin >> n >> k >> s;
  vector<int> r(1);
  for(int i = 1; i < n; i++)
    if(s[i - 1] != s[i]) r.push_back(i);
  r.push_back(n);
  for(int i = 0; i < r.size() - 1; i++) {
    if(s[r[i]] == '0')
      m = max(m, r[min(i + k * 2, (int)r.size() - 1)] - r[i]);
    else
      m = max(m, r[min(i + k * 2 + 1, (int)r.size() - 1)] - r[i]);
  }
  cout << m << '\n';
}