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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int main()
{
  int64_t N, K, ans = 0;
  string S;
  cin >> N >> K >> S;
  vector<int64_t> Count(1, 0);
  S.at(0) == '0' ? Count[0]-- : Count[0]++; // 0なら-1, iなら+1
  for (int i = 1; i < N; i++)
  {
    if (S.at(i) != S.at(i - 1))
      Count.push_back(0);
    S.at(i) == '0' ? Count.at(Count.size() - 1)-- : Count.at(Count.size() - 1)++;
  }

  for (int i = 0; i < Count.size(); i++)
    ans = max(ans, Count.at(i));

  vector<int64_t> Sum(Count.size() + 1, 0);
  for (int i = 0; i < Count.size(); i++)
    Sum.at(i + 1) = Sum.at(i) + abs(Count.at(i));

  int64_t length = Count.size(), zero = 0;
  for (int64_t i = 0; i < length; i++)
  {
    if ((K % 2 == 1 && Count.at(i) > 0) || (K % 2 == 0 && Count.at(i) < 0))
      continue;
    int64_t sum = 0;
    int64_t r = min(length - 1, i + K) + 1;
    int64_t l = max(zero, i - K);
    sum += Sum.at(r) - Sum.at(l);
    ans = max(ans, sum);
  }
  cout << ans << endl;
}