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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
//TLE
struct st{
  char a;
  int b;
};

int min(int a, int b) {
  if (a < b) return a;
  else return b;
}

int main() {
  int n, k;
  scanf("%d %d\n", &n, &k);
  struct st st[n];
  char s, t;
  scanf("%c", &s);
  t = s;
  st[0].a = s;
  st[0].b = 1;
  int j = 0;
  for (int i = 1; i < n; i++) {
    scanf("%c", &s);
    if (s == t) {
      st[j].b += 1;
    } else {
      j++;
      st[j].a = s;
      st[j].b = 1;
      t = s;
    }
  }
  n = j+1;
  int count = 0, max = 0;
  if (n <= 2*k-1) {
    for (int i = 0; i < n; i++) {
      count += st[i].b;
    }
    printf("%d\n", count);
    return 0;
  }
  for (int i = 0; i < n; i++) {
    count = 0;
    if (st[i].a == '0') {
      for (int j = 0; j < min(2*k, n-i); j++) {
        count += st[i+j].b;
      }
    } else {
      for (int j = 0; j < min(2*k+1, n-i); j++) {
        count += st[i+j].b;
      }
    }
    if (count > max) {
      max = count;
    }
  }
  printf("%d\n", max);
  return 0;
}