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 <stdio.h>
#include <string.h>
#include <vector>
using namespace std;
typedef long long ll;

int main(void) {
  ll i, j, k, n, cnt = 0, ans = 0, v = 0;
  scanf("%lld%lld", &n, &k);
  char s[n + 10];
  vector<ll> vec;
  scanf("%s", s);
  vec.push_back(0);
  if(s[0] == '0') cnt++;
  for(i = 1; i < n; ++i) {
    if(s[i] == '0') {
      if(s[i - 1] == '1') {
        if(ans < i  - vec[v]) ans = i - vec[v];
        cnt++;
        if(cnt > k) v++;
      }
    } else {
      if(s[i - 1] == '0') vec.push_back(i);
    }
  }
  if(ans < n - vec[v]) ans = n - vec[v];
  printf("%lld", ans);
  return 0;
}