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())
li = [int(i) for i in list(input())]
kukanidx = []
idx = 0
for i in li:
    if i != li[idx-1]:
        kukanidx.append(idx)
    idx += 1
idx = 1
kukanidx.append(n)

kukan = kukanidx.copy()
for i in kukanidx[1:]:
    kukan[idx] = kukanidx[idx]-kukanidx[idx-1]
    idx += 1

anskouho = []
idx = 0
if li[0] == 1:
    for i in kukan:
        anskouho.append(sum(kukan[idx:idx+(k*2+1)]))
        idx += 1
        if idx > len(kukan)-(k*2+1):
            break
else:
    for i in kukan:
        anskouho.append(sum(kukan[idx:idx+(k*2)]))
        idx += 1
        if idx > len(kukan)-(k*2):
            break

print(max(anskouho))