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(){
  int n, k; cin >> n >> k;
  string s; cin >> s;
  vector<int> accum(n, 0);
  int acc = 0;
  for(int i = 0; i < s.size(); i++){
    if(s[i] == '0' && (i == 0 || s[i - 1] != '0')) accum[i] = 1;
  }
  for(int i = 1; i < s.size(); i++) accum[i] += accum[i - 1];
  int lo = 1, hi = s.size(), ans = 1;
  while(lo <= hi){
    int mid = (lo + hi)/2;
    bool someone = false;
    for(int i = 0; i+mid <= s.size(); i++){
    	bool canDo = s[i] == '1' || (s[i] == '0' && (i == 0 || s[i - 1] != '0'));
      if(canDo){
        int V = accum[i + mid] - (i ? accum[i - 1] : 0);
        if(V <= k) someone = true;
      }
    }
    if(someone) lo = mid + 1, ans = mid;
    else hi = mid - 1;
  }
  cout << ans << endl;
  return 0;
}