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; cin >> n >> k;
  string s; cin >> s;
  vector<int> hoge = {0,1};
  char tmp = s.at(0);
  for (int i=1; i<n; i++) {
    if (s.at(i) == tmp) hoge.back()++;
    else {
      tmp = s.at(i);
      hoge.push_back(hoge.back()+1);
    }
  }
  if (s.at(0) == '0') hoge.insert(hoge.begin(), 0);
  if (s.back() == '0') hoge.push_back(hoge.back());

  int ans = 0;
  if (hoge.size() > 2*k+1) {
    for (int i=0; i<hoge.size()-(2*k+1); i+=2) ans = max(ans, hoge.at(i+2*k+1) - hoge.at(i));
  }
  else ans = hoge.back();
  cout << ans << endl;
}