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;
typedef long long lint;
typedef pair<int, int> P;

int main(){
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int N, K, cnt = 0, right = 0, ans = 0; cin >> N >> K;
    string S; cin >> S;
    for(int i=0;i<N;i++){
        while(right<N and (cnt<K or S[right]=='1')){
            if(S[right]=='0'){
                cnt++;
                while(right<N and S[right]=='0') right++;
            }
            else right++;
        }
        ans = max(ans, right-i);

        if(i<N-1 and S[i]=='0' and S[i+1]=='1') cnt--;
    }
    cout << ans << endl;
    return 0;
}