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>

#define MAX_N 100010

using namespace std;

int main() {
    int n, k;
    scanf("%d %d\n", &n, &k);
    int j = 0, ans = 0, change[MAX_N];
    char prev = '0';
    for (int i=0; i<n; i++) {
        char c;
        scanf("%c", &c);
        if (c != prev) {
            prev = c;
            change[j] = i;
            j++;
        }
        if (c == '1')
            ans = max(i+1 - (j>=2*k+1 ? change[j-2*k-1] : 0), ans);
        else if (k > 0)
            ans = max(i+1 - (j>=2*k ? change[j-2*k] : 0), ans);
    }
    printf("%d\n", ans);
    return 0;
}


