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 <iostream>
#include <vector>

using namespace std;
using vi = vector<int>;

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define IN(v) int v;cin>>v;
#define OUT(v) cout<<v<<"\n"

int main() {
  cin.tie(0); ios::sync_with_stdio(false);
  IN(N); IN(K);

  string S;
  cin >> S;

  int before = 1,
      count = 0;
  vi v;
  for (const auto& c : S) {
    if ((c == '0' && before == 0) || (c == '1' && before == 1))
      ++count;
    else {
      v.push_back(count);
      before = c - '0';
      count = 1;
    }
  }
  v.push_back(count);

  int ans = 0;
  REP(i, v.size()) {
    if (i % 2 == 1)
      continue;
    int tmp = 0;
    REP(j, 2 * K + 1) {
      if (i + j >= v.size())
        break;
      tmp += v[i + j];
    }
    ans = max(ans, tmp);
  }
  OUT(ans);
  return 0;
}