#include <stdio.h>
typedef long long ll;

int main(void) {
  ll i, j, k, n, c;
  scanf("%lld%lld%lld", &n, &k, &c);
  char s[n + 10];
  scanf("%s", s);
  ll cnt[2][n], las = -1e9;
  cnt[0][0] = cnt[1][n - 1] = 0;
  for(i = 1; i < n; ++i) {
    cnt[0][i] = cnt[0][i - 1];
    if(s[i - 1] == 'o' && i - 1 > las + c) cnt[0][i]++, las = i - 1;
  }
  las = 1e9;
  for(i = n - 2; i >= 0; --i) {
    cnt[1][i] = cnt[1][i + 1];
    if(s[i + 1] == 'o' && i + 1 < las - c) cnt[1][i]++, las = i + 1;
  }
  for(i = 0; i < n; ++i) if(cnt[0][i] + cnt[1][i] < k) printf("%lld\n", i + 1);
  return 0;
}
### Prompt

Your task is to create a CPP solution to the following problem:

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.

Examples

Input

11 3 2
ooxxxoxxxoo


Output

6


Input

5 2 3
ooxoo


Output

1
5


Input

5 1 0
ooooo


Output




Input

16 4 3
ooxxoxoxxxoxoxxo


Output

11
16

### Response

```cpp
#include <stdio.h>
typedef long long ll;

int main(void) {
  ll i, j, k, n, c;
  scanf("%lld%lld%lld", &n, &k, &c);
  char s[n + 10];
  scanf("%s", s);
  ll cnt[2][n], las = -1e9;
  cnt[0][0] = cnt[1][n - 1] = 0;
  for(i = 1; i < n; ++i) {
    cnt[0][i] = cnt[0][i - 1];
    if(s[i - 1] == 'o' && i - 1 > las + c) cnt[0][i]++, las = i - 1;
  }
  las = 1e9;
  for(i = n - 2; i >= 0; --i) {
    cnt[1][i] = cnt[1][i + 1];
    if(s[i + 1] == 'o' && i + 1 < las - c) cnt[1][i]++, las = i + 1;
  }
  for(i = 0; i < n; ++i) if(cnt[0][i] + cnt[1][i] < k) printf("%lld\n", i + 1);
  return 0;
}
```