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: Runtime Error)

#include <iostream>
#include <vector>
#include <string>
#include <queue>
#include <algorithm>
#define MOD 1000000007
#define INF 1 << 30

using namespace std;
typedef long long ll;

int main(void) {
	int num, i, k, len = 0, j = 0, a, ans = 0, sum = 0;
	string s;
	queue<int> que;
	cin >> num >> k >> s;
	vector<int> d(num + 2, 0);
	if (s[0] == '0')
		len++;
	for (i = 0; i < num;) {
		a = 0;
		while (i < num && s[i + a] == s[i])
			a++;
		i += a;
		if (a > 0)
			d[len++] = a;
	}
	if (s[num - 1] == '0')
		len++;
	k = (k << 1) + 1;
	for (i = 0; i < k; i++) {
		que.push(d[i]);
		sum += d[i];
	}
	ans = sum;
	for (; i < len; i += 2) {
		sum += d[i] + d[i + 1];
		sum -= que.front();
		que.pop();
		sum -= que.front();
		que.pop();
		ans = max(ans, sum);
	}
	cout << ans << "\n";
	return 0;
}