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 n, k, seq[100005], t, ans, tmp;
char s[100005];
int main(){
    scanf("%d%d%s", &n, &k, s);
    if(s[0] == '0') t++;
    for(int i = 0; i < n; i++){
        if(t%2 == s[i] - '0') t++;
        seq[t]++;
    }
    if(s[n-1] == '0') t++;
    t++;
    if(t <= k*2+1) return 0*printf("%d\n", n);
    for(int i = 0; i <= k*2; i++) tmp += seq[i];
    ans = tmp;
    for(int i = k*2; i < t-2; i += 2){
        tmp += seq[i+1] + seq[i+2] - seq[i-k*2] - seq[i-k*2+1];
        ans = max(ans, tmp);
    }
    printf("%d\n", ans);
}