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 = input()
n_k_list = n_k.split()
K = int(n_k_list[1])
people = input()
people_list = []
now = people[0]
for i in people[1:]:
    if now[-1] == i:
        now += i
    else:
        people_list.append(now)
        now = i
people_list.append(now)

max_count = 0
change_index = None
while K:
    for index, cell in enumerate(people_list):
        if cell[0] == "0":
            if index == 0 and index == people_list.__len__() - 1:
                this_count = cell.__len__()
            elif index == 0:
                this_count = cell.__len__() + people_list[index + 1].__len__()
            elif index == people_list.__len__() - 1:
                this_count = cell.__len__() + people_list[index - 1].__len__()
            else:
                this_count = cell.__len__() + people_list[index - 1].__len__() + people_list[index + 1].__len__()
            if max_count < this_count:
                max_count = this_count
                change_index = index
    if change_index is not None:
        change_char = ""
        for z in range(people_list[change_index].__len__()):
            change_char += "1"
        if change_index == 0 and change_index == people_list.__len__() - 1:
            new_char = change_char
            people_list[change_index] = new_char
        elif change_index == 0:
            new_char = people_list[change_index + 1] + change_char
            people_list[change_index] = new_char
            del people_list[change_index + 1]
        elif change_index == people_list.__len__() - 1:
            new_char = people_list[change_index - 1] + change_char
            people_list[change_index] = new_char
            del people_list[change_index - 1]
        else:
            new_char = people_list[change_index - 1] + people_list[change_index + 1] + change_char
            people_list[change_index] = new_char
            del people_list[change_index + 1]
            del people_list[change_index - 1]
        change_index = None
        K -= 1
    else:
        break
print(max(people_list, key=len).__len__())