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;
  S += '/';
  for (int i = 0, cnt = 1; i + 1 < S.size(); i++) {
    if (S.at(i) == S.at(i + 1)) cnt++;
    else {
      V.push_back(cnt);
      cnt = 1;
    }
  }
  if (S.front() != '1') V.insert(V.begin(), 0);

  N = V.size();
  vector<int> C(N + 1);
  for (int i = 1, sum = 0; i < N + 1; i++) {
    C.at(i) = sum += V.at(i - 1);
  }

  int mx = 0;
  for (int i = 0; i < N + 1; i += 2) {
    mx = max(mx, C.at(min(i + K * 2 + 1, N)) - C.at(i));
  }
  cout << mx << "\n";
}