Problem:
Score : 500 points

Problem Statement

Takahashi has decided to work on K days of his choice from the N days starting with tomorrow.

You are given an integer C and a string S. Takahashi will choose his workdays as follows:

After working for a day, he will refrain from working on the subsequent C days.

If the i-th character of S is x, he will not work on Day i, where Day 1 is tomorrow, Day 2 is the day after tomorrow, and so on.

Find all days on which Takahashi is bound to work.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq K \leq N

0 \leq C \leq N

The length of S is N.

Each character of S is o or x.

Takahashi can choose his workdays so that the conditions in Problem Statement are satisfied.

Input

Input is given from Standard Input in the following format:

N K C
S

Output

Print all days on which Takahashi is bound to work in ascending order, one per line.

Sample Input 1

11 3 2
ooxxxoxxxoo

Sample Output 1

6

Takahashi is going to work on 3 days out of the 11 days. After working for a day, he will refrain from working on the subsequent 2 days.

There are four possible choices for his workdays: Day 1,6,10, Day 1,6,11, Day 2,6,10, and Day 2,6,11.

Thus, he is bound to work on Day 6.

Sample Input 2

5 2 3
ooxoo

Sample Output 2

1
5

There is only one possible choice for his workdays: Day 1,5.

Sample Input 3

5 1 0
ooooo

Sample Output 3

There may be no days on which he is bound to work.

Sample Input 4

16 4 3
ooxxoxoxxxoxoxxo

Sample Output 4

11
16

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

int main() {
  int n, k, c;
  cin >> n >> k >> c;
  string s;
  cin >> s;

  if (c == 0 || k == 1) {
    puts("");
    return 0;
  }

  vector<int> left(n, -1);
  vector<int> right(n, -1);

  int rest = -1;
  int cur = -1;
  for (int i = 0; i < n; ++i) {
    rest--;
    if (rest < 0 && s[i] == 'o') {
      left[i] = ++cur;
      rest = c;
    }
  }

  rest = -1;
  cur = -1;
  for (int i = n - 1; i >= 0; --i) {
    rest--;
    if (rest < 0 && s[i] == 'o') {
      right[i] = ++cur;
      rest = c;
    }
  }

  for (int i = 0; i < n; ++i) {
    if (right[i] >= 0) right[i] = cur - right[i];
  }

  for (int i = 0; i < n; ++i) {
    if (right[i] == left[i] && right[i] >= 0) {
      cout << i + 1 << '\n';
    }
  }
  return 0;
}