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()
Nums = []
now = 1 # 今見ている数
cnt = 0 # nowがいくつ並んでいるか
for i in range(N):
    if S[i] == str(now):
        cnt += 1
    else:
        Nums.append(cnt)
        now = 1 - now   # 0と1を切り替えるときの計算 now ^= 1
        cnt = 1
if cnt != 0:
    Nums.append(cnt)
# 1-0-1-0-1-0-1 って感じの配列が欲しい
# 1-0-1-0-1-0 みたいに0で終わっていたら適当に1つ足す
if len(Nums) % 2 == 0:
    Nums.append(0)

Add = 2 * K + 1
ans = 0
# 尺取り法 forループの外側にleft, rightを持つ
left = 0
right = 0
tmp = 0 # [left, right)のsum
# 1-0-1... の1から始めるので、偶数番目だけ見る
for i in range(0, len(Nums), 2):
    # 次のleft, rightを計算する
    Nextleft = i
    Nextright = min(i + Add, len(Nums))
    # 左端を移動する
    while Nextleft > left:
        tmp -= Nums[left]
        left += 1
    # 右端を移動する
    while Nextright > right:
        tmp += Nums[right]
        right += 1
    ans = max(tmp, ans)

print(ans)