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 <iostream>
#include <string>
#include <cstring>
using namespace std;

size_t n;
int k, m;
string s;
size_t j[100010];
bool f;
int x, t;

int main(){
    cin >> n >> k;
    cin >> s;

    m = 1;
    j[0] = 0;
    for (size_t i = 1; i < s.size(); ++i){
        if (s[i] != s[i-1]){
            j[m] = i;
            m++;
        }
    }
    j[m] = n;

    x = 0;
    for (int i = 0; i < m; ++i){
        if (s[j[i]] == '0') t = j[min(m, i+2*k)] - j[i];
        else t = j[min(m, i+2*k+1)] - j[i];
        if (x < t) x = t;
    }

    cout << x;
}