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: Wrong Answer)

n,k = list(map(int, input().split()))
s = input()
s += '0' if s[n-1] == '1' else '1'
c = []

j = 0
for i in range(1, n+1):
	if s[i-1] == s[i]:
		j += 1
	else:
		c.append([s[i-1], j+1])
		j = 0

for i in range(1, len(c)):
	c[i][1] += c[i-1][1]

if 2*k >= len(c):
	print(c[-1][1])
	quit()
else:
	p = 0
	res = 0
	if s[0][0] == '0':
		res = c[1][1] if k == 1 else c[2*k+1][1]
		p += 1
	else:
		res = c[2][1] if k == 1 else c[2*k][1]
		p += 2

	while p+2*k < len(c):
		res = max(res, c[p+2*k][1] - c[p-1][1])
		p += 2

	res = max(res, c[-1][1] - c[(len(c)-1) - 2*k][1]) if c[-1][0] == '0' else max(res, c[-1][1] - c[(len(c)-1) - (2*k+1)][1])

	print(res)
	quit()