Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

5 1
00010

Sample Output 1

4

We can have four consecutive people standing on hands, which is the maximum result, by giving the following direction:

Give the direction with l = 1, r = 3, which flips the first, second and third persons from the left.

Sample Input 2

14 2
11101010110011

Sample Output 2

8

Sample Input 3

1 1
1

Sample Output 3

1

No directions are necessary.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#define ll long long
#define ull unsigned long long

using namespace std;

int main() {
	int N, K;
	cin >> N >> K;
	string S;
	cin >> S;

	int ans = 0;
	int r = 0;
	int cnt = 0;
	if (S[r]=='0'){
		cnt++;
	}
	for (int l = 0; l < N; ++l){
		//cntがKより大きくなるまで右端を動かす
		while( (cnt <= K) && (r < N-1) ) {
			if ( (S[r]=='1') && (S[r+1]=='0') ){
				cnt++;
			}
			if (cnt > K) break;
			r++;
			if (r==N-1) break;
		}
		int tmp = r - l + 1;
		ans = max(ans, tmp);

		// printf("tmp=%i, ans=%i at (l,r)=(%i,%i)\n", tmp, ans, l, r);

		if (l < N-1){
			if ( (S[l]=='0') && (S[l+1]=='1') ){
				// printf("cnt %i -> ", cnt);
				cnt--;
				if (r < N-1){
					r++;
				}
				// printf("%i\n", cnt);
			}
		}
	}

	cout << ans << "\n";


	return 0;
}