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;
using ll = long long;

int main() {
    int N, K;cin >> N >> K;
    string S;cin >> S;

    vector<int> X;
    if (S[0] == '0') X.push_back(0);
    for (int i = 0;i < N;){
        int j = i;
        while (j < N && S[i] == S[j]) j++;
        X.push_back(j - i);
        i = j;
    }
    vector<int> Y(X.size() + 1,0);
    for (int i = 0;i < X.size();i++) Y[i + 1] = Y[i] + X[i];

    int res = 0;
    for (int i = 0;i < Y.size();i += 2){
        int j = i + K * 2 + 1;
        if (j >= Y.size()) j = Y.size() - 1;
        res = max(res,Y[j] - Y[i]);
    }
    cout << res << endl;
}