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: Accepted)

from itertools import groupby
n, k = map(int, input().split())
s = input()

t = []
if s[0] == "0":
    t.append(("1", 0))
for c, v in groupby(s):
    t.append((c, sum(1 for _ in v)))

if s[-1] == "0":
    t.append(("1", 0))

S = [t[0][1]]
rm = [0]
m = len(t)
for i in range(m//2):
    tmp = S[-1]
    S.append(tmp+t[2*i+1][1]+t[2*i+2][1])
    rm.append(tmp+t[2*i+1][1])

ans = []
for i in range(m//2+1):
    if i - k < 0:
        ans.append(S[i])
    else:
        ans.append(S[i]-rm[i-k])
print(max(ans))