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; 
int main(){
	int N, K; 
	cin >> N >> K; 
	string S; 
	cin >> S; 
	vector<int> L = {}; 
	vector<int> R = {}; 
	R.push_back(0);
	for(int i = 0; i < N; i++){
		if((i == 0 && S.at(i) == '0') || (i != 0 && (S.at(i-1) == '1' && S.at(i) == '0'))) 
			L.push_back(i+1); 
		if((i == N - 1 && S.at(i) == '0') || (i != N- 1 && (S.at(i) == '0' && S.at(i+1) == '1')))
			R.push_back(i+1);  
	}
	L.push_back(N+1); 
	int ans = 0; 
	int M = L.size(); 
	if(K >= M - 1) ans = N; 
	else{
		for(int i = 0; i < M - K; i++){
			ans = max(ans, L[i+K] - R[i] - 1); 
		}
	}
	cout << ans << endl; 
}