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: Time Limit Exceeded)

#include <algorithm>
#include <iostream>
#include <map>
#include <numeric>
#include <set>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <vector>
#include <queue>
#define MOD (1000000007l)
#define rep(i, n) for(long i = 0; i < n; i++)

using namespace std;

int main(void) {
  long N, K;
  cin >> N >> K;
  string S;
  cin >> S;
  char current = S[0];
  long count = 1;
  vector<long> foo;
  for (long i = 1; i < S.size(); i++) {
    if (S[i] == current) count++;
    else {
      foo.push_back(count * (current == '0' ? -1 : 1));
      count = 1;
      current = S[i];
    }
  }
  foo.push_back(count * (current == '0' ? -1 : 1));
  cerr << S << endl;
  cerr << "-----" << endl;
  for (auto it: foo) cerr << it << endl;
  cerr << "-----" << endl;
  long result = 0;
  for (long i = 0; i < foo.size(); i++) {
    long can = K;
    long tmp = 0;
    for (long j = i; j < foo.size(); j++) {
      if (foo[j] < 0 and can <= 0) break;
      if (foo[j] > 0) tmp += foo[j];
      else {
        tmp -= foo[j];
        can--;
      }
    } 
    if (tmp > result) result = tmp;
  }
  cout << result << endl;
  return 0;
}