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: Accepted)

#include <bits/stdc++.h>
#define REP(i,n) for(int i=0;i<(n);++i)
using namespace std;
typedef long long ll;

int main() {
  int N, K;
  string S;
  cin >> N >> K >> S;

  if (K>=N) {
    cout << N << endl;
    return 0;
  }

  vector<int> C,R;
  R.push_back(0);
  if (S[0]=='0' ) C.push_back(0);
  S = S + 'Z';
  int cnt = 1;
  REP(i,N){
    if ( S[i] != S[i+1] ) {
      C.push_back(cnt);
      cnt = 1;
    } else {
      cnt++;
    }
  }
  if (S[N-1]=='0' ) C.push_back(0);
  
  REP(i, C.size())
    R.push_back( R[i]+C[i] );

  int smax = 0;
  for ( int l=0; (2*l)<R.size(); ++l ) {
    int ridx = min(2*(l+K-1)+3, (int)R.size()-1);
    smax = max( smax, R[ridx] - R[2*l] );
//    cout << "- [" << l << "," << (2*(l+K-1)+3) << "] " << R[2*(l+K-1)+3] << " - " << R[2*l] << endl;
  }
//  REP(i,R.size())
//    cout << C[i] << " " << R[i] <<endl;
  cout << smax << endl;
}