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 <iostream>
#include <vector>
using namespace std;
int main(){
  int n,k;cin>>n>>k;
  string s;cin>>s;
  int begi = s[0]-'0';
  vector<int> assy;
  int inp = 1;
  for(int i = 1; s.size() > i; i++){
    if(s[i] != s[i-1]){
      assy.push_back(inp);
      inp = 0;
    }
    inp++;
  }
  assy.push_back(inp);
  if((begi == 0 ? assy.size() % 2 + assy.size()/2 : assy.size()/2) <= k){
    cout << n << endl;
    return 0;
  }
  long long mawasi = k*2;
  if(begi == 1)mawasi++;
  int syoki = 0;
  for(int i = 0; mawasi > i; i++){
    syoki += assy[i];
  }
  int mx = syoki;
  for(int i = mawasi; assy.size() > i; i+=2){
    if(i == mawasi && begi == 0){
      syoki -= assy[i-mawasi];
    }else{
      syoki -= assy[i-mawasi]+assy[i-mawasi+1];
    }
    syoki += assy[i];
    if(i+1 != n)syoki += assy[i+1];
    mx = max(mx,syoki);
  }
  cout << mx << endl;
}