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<algorithm>
#include<vector>
using namespace std;

int main(){
	int N,K;
	cin>>N>>K;
	vector<int> count;
	count.push_back(0);
	char bc='0';
	for(int i=0;i<N;++i){
		char ic;
		cin>>ic;
		if(bc == ic){
			++count.back();
		}else{
			count.push_back(1);	
			bc = ic;
		}
	}
	int ans=0;
	int nc = 0;
	int lp=0;
	int zc=0;
	for(int i=0;i<count.size();++i){
		nc += count[i];
		if(i%2 == 0){
			++zc;
			if(zc > K){
				nc -= count[lp];
				++lp;
				if(lp%2 == 0){
					nc -= count[lp];
					++lp;
				}
				--zc;
			}
		}
		ans = max(ans, nc);
	}
	cout << ans << endl;
	return 0;
}
