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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

n,k = map(int,input().split())
s = input()
x = len(s)
lst = [0]
ans = 0
for i in range(1,x):
    if s[i-1] != s[i]:
        lst.append(i)
lst.append(x)
#if s[0] == "1" and s[x-1] == "0":
    #s.reverse()
if s[0] == "0" and s[x-1] == "1":
    if 2*k >= len(lst):
        ans = x
    else:
        i = 1
        while i+2*k+1 <= len(lst)-1:
            ##ans = max(ans,lst[i+2*k+1]-lst[i])
            i += 2
        #ans = max(ans,lst[2*k])
if s[0] == "0" and s[x-1] == "0":
    if 2*k >= len(lst):
        ans = x
    else:
        i = 1
        while i+2*k+1 <= len(lst)-1:
            ##ans = max(ans,lst[i+2*k+1]-lst[i])
            i += 2
        #ans = max(ans,lst[2*k])
        #ans = max(ans,x-lst[len(lst)-2*k-1])
if s[0] == "1" and s[x-1] == "1":
    if 2*k+1 >= len(lst):
        ans = x
    else:
        i = 0
        while i+2*k+1 <= len(lst)-1:
            #ans = max(ans,lst[i+2*k+1]-lst[i])
            i += 2
print(ans)