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()

a = []

before = s[0]
first = int(s[0])
count = 1
for x in s[1:]:
    if x != before:
        a.append(count)
        count = 1
    else:
        count += 1
    before = x
a.append(count)

if len(a) < k * 2:
    print(n)
    exit()

ans = 0
for i in range(len(a) - (k * 2) + 1):
    temp = 0
    mode = first
    x = k
    for p in range(i, len(a)):
        #print(i, mode, temp, x)
        if mode == 0:
            if x == 0:
                break
            x -= 1
        temp += a[p]
        p += 1
        mode = (mode + 1) % 2
    ans = max(temp, ans)

print(ans)