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;

#define f(i,a,b) for (int i = a; i < b; i++)
#define pb push_back

int main() {
int n,k; cin>>n>>k;
string s; cin>>s;
vector<pair<int,int>> res;
f(i,0,n) {
int j = i;
while (j < n && s[i] == s[j]) j++;
res.pb({j-i,s[i]-'0'});
i = j-1;
}
int sum = 0, num = 0, j = 0;
int ans = 0;
f(i,0,res.size()) {
while (j < res.size() && num+(res[j].second==0) <= k) {
num += res[j].second == 0;
sum += res[j].first;
j++;
}
ans = max(ans,sum);
if (res[i].second == 0) num--;
sum -= res[i].first;
}
cout << ans << endl;
return 0;
}