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 = map(int, input().split())                                                                               
s = input()

def count_consecutive(s):
    return max(list(map(lambda x: len(x), s.split('0'))))

def f(s, k):
    global mx
    for j in range(len(pos)-k+1):
        tmp = s
        for n in range(k):
            i, count = pos[j+n]
            tmp = tmp[:i] + '1'*count + tmp[i+count:]
        mx = max(count_consecutive(tmp), mx)
    return mx

l, mx, pos = len(s), count_consecutive(s), []
i = 0 
while i < l:
    if s[i] == '0':
        count = 0 
        while i+count < l and s[i+count] == '0':
            count += 1
        pos.append([i, count])
        i += count
    else:
        i += 1
print(f(s, k))