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 Python (3.4.3) (judge verdict: Accepted)

n, k = map(int, input().split())
s = input()
s = list(map(int, s))

v = []
cnt = 1
for i in range(n - 1):
	if s[i] == s[i + 1]:
		cnt = cnt + 1
	else:
		v.append(cnt)
		cnt = 1
v.append(cnt)
m = len(v)

sum = [0] * (m + 1)
for i in range(1, m + 1):
	sum[i] = sum[i - 1] + v[i - 1]

ans = 0
for i in range(m + 1):
	if s[0] == 0:
		if i % 2 == 0:
			ans = max(ans, sum[i] - sum[max(i - k * 2 - 1, 0)])
		else:
			ans = max(ans, sum[i] - sum[max(i - k * 2, 0)])
	else:
		if i % 2 == 0:
			ans = max(ans, sum[i] - sum[max(i - k * 2, 0)])
		else:
			ans = max(ans, sum[i] - sum[max(i - k * 2 - 1, 0)])

print(ans)