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: Accepted)

n,k = map(int,input().split())
s = input()
s = s + '#'
left = 0
right =0
l = []
while left < n:
    if s[right] == s[right+1]:
        right += 1
    else:
        l.append(right-left+1)
        right += 1
        left = right
if s[0] == '0':
    l.insert(0,0)
if s[n-1] == '0':
    l.append(0)
n = len(l)
l_sum = [0]*(n)
l_sum[0] = l[0]
for i in range(1,n):
    l_sum[i] = l[i] + l_sum[i-1]
if n <= 2*k + 1:
    print(l_sum[n-1])
else:
    res = l_sum[2*k]
    for i in range(2,n-2*k,2):
        res = max(res,l_sum[i+2*k]-l_sum[i-1])
    print(res)