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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

int main(void)
{
	int num_of_human;
	int limit_of_instruction;
	scanf("%d%d", &num_of_human, &limit_of_instruction);

	char init_state[num_of_human + 1];
	scanf("%s", init_state);

	int compact_expression[num_of_human];
	int i, j, k, sum;
	j = 0;
	for(i = 0; i < num_of_human; i++) {
		sum = 0;

		if(j >= num_of_human) {
			while(i < num_of_human) {
				compact_expression[i] = 0;
				i++;
			}
			break;
		}

		if(i % 2 == 0) {
			while(init_state[j] == '1') {
				j++;
				sum++;
			}
			compact_expression[i] = sum;
		} else {
			while(init_state[j] == '0') {
				j++;
				sum++;
			}
			compact_expression[i] = sum;
		}
	}

	int ans;

	if(limit_of_instruction * 2 >= i) {
		printf("%d\n", num_of_human);
		return 0;
	}

	ans = 0;
	for(j = 0; j + limit_of_instruction * 2 < i; j += 2) {
		sum = 0;
		for(k = 0; k < limit_of_instruction * 2 + 1; k++) {
			sum += compact_expression[j + k];
		}
		if(ans < sum)
			ans = sum;
	}

	printf("%d\n", ans);

	return 0;
}