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 pair<int,char> P;

int main() {
    int N,K,i,m=0,t,l;
    string S;
    vector<int> h;
    cin >> N >> K >> S;
    char p=S.at(0);

    h.push_back(0);
    for(i=1; i<N; i++) {
        if(p != S.at(i))
            h.push_back(i);
        p = S.at(i);
    }
    h.push_back(N);

    l = h.size()-1;
    for(i=0; i<h.size()-1; i++) {
        if(S.at(h.at(i)) == '0')
            t = h.at(min(i+K*2, l));
        else
            t = h.at(min(i+K*2+1, l));
        m = max(m, t-h.at(i));
    }

    cout << m << endl;
}