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;
#define endl "\n"

int main(void) {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    int n, k;
    cin >> n >> k;
    cin.ignore();
    string str;
    cin >> str;
    int i = 0, j = 0, ans = 0;
    while (j < n) {
        while (j < n && str[j] == '1')  j++;
        ans = max(ans, j - i);
        while (j < n && str[j] == '0')  j++;
        --k;
        if (k < 0) {
            while (i < n && str[i] == '1')  i++;
            while (i < n && str[i] == '0')  i++;
        }
        ans = max(ans, j - i);
    }
    cout << ans << endl;
}
