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 <cstdio>
#include <algorithm>

int cum[100001];

int main()
{
  int n, k;
  char s[100001];
  scanf("%d%d%s", &n, &k, s);
  int blk[100000];
  int* dst = blk;
  for (char *p = s; *p;)
  {
    char* q = p;
    while (*p && *p == *q) ++p;
    *dst++ = p - q;
  }
  int nblk = dst - blk;
  for (int i = 0; i < nblk; ++i)
  {
    cum[i + 1] = cum[i] + blk[i];
  }

  int ans = 0;
  if (*s == '0')
  {
    ans = cum[std::min(2 * k, nblk)];
  }
  for (int i = ('1' - *s); i < nblk; i += 2)
  {
    ans = std::max(ans, cum[std::min(i + 2 * k + 1, nblk)] - cum[i]);
  }
  printf("%d\n", ans);
  return 0;
}