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 ll;

int main() {
    int n,k;string s;
    cin>>n>>k>>s;
    vector<int> a;
    a.push_back(0);
    int c=1;
    for (int i=0;i<n;i++) {
        if (i==n-1)
            a.push_back(a.back()+c);
        else if (s[i]==s[i+1])
            c++;
        else {
            a.push_back(a.back()+c);
            c=1;
        }
    }
    int mx=0;
    if (s[0]=='0')
        mx=max(mx,a[min(2*k,(int)a.size()-1)]-a[0]);
    for (int i=s[0]=='0';i<a.size();i+=2)
        mx=max(mx,a[min(i+2*k+1,(int)a.size()-1)]-a[i]);
    cout<<mx<<endl;
    return 0;
}