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()

ns = []
now_num = s[0]
cnt = 0
for i in range(n):
    if now_num == s[i]:
        cnt += 1
    else:
        ns.append((now_num, cnt))
        now_num = s[i]
        cnt = 1
ns.append((now_num, cnt))
len_ns = len(ns)
pre = [0] * (len_ns + 1)
for i in range(len_ns):
    pre[i + 1] = pre[i] + ns[i][1]

ans = 1
for i in range(len_ns - 2*k + 1):
    temp = pre[i + 2*k] - pre[i]
    if i > 0 and ns[i - 1][0] == '1':temp += ns[i - 1][1]
    if i + 2*k < len_ns and ns[i + 2*k][0] == '1':temp += ns[i + 2*k][1]
    if temp > ans:ans = temp
print(ans if len_ns - 2*k + 1 > 0 else len(s))