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 = list(input())

if s[0] == "0":
    s_num = [0]
else:
    s_num = []
pre = s[0]
num = 0

for i in s:
    if pre == i:
        num += 1
    else:
        s_num.append(num)
        num = 1
        pre = "1" if pre == "0" else "0"
else:
    s_num.append(num)

sums = [s_num[0]]

for i,e in enumerate(s_num[1:]):
    sums.append(sums[i] + e)

ret = 0
sums_len = len(sums)-1
for i in range(0,len(s_num)//2+1,2):
    ret = max(ret,(sums[min(i+k*2+1,sums_len)] - sums[i]))
 
print(ret)