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: Time Limit Exceeded)

n,k = map(int,input().split(' '))
s = input()
l = []
l_last = 1
count = 0
for s0 in s:
    if int(s0) == l_last:
        count += 1
    else:
        l.append(count)
        count = 1
        l_last = (l_last+1)%2
if count!=0:
    l.append(count)
if len(l)%2==0:
    l.append(0)
q=(len(l)-1)/2
if k >= q:
    ans = n
else:
    ans = 0
    j = 0
    while j <= q-k:
        if ans <= sum(l[2*j:2*(j+k)+1]):
            ans = sum(l[2*j:2*(j+k)+1])
        j += 1
print(ans)