#include <stdio.h>
using namespace std;

int N, K, C;
char S[200200]; int P[200200], Q[200200];

int main()
{
	scanf ("%d %d %d", &N, &K, &C);
	scanf ("%s", S);

	for (int i = 0, c = 0; i < N; i++) if (S[i] == 'o'){
		P[i] = ++c;
		i += C;
		if (c == K) break;
	}

	for (int i = N - 1, c = K; i >= 0; i--) if (S[i] == 'o'){
		Q[i] = c--;
		i -= C;
		if (c == 0) break;
	}

	for (int i = 0; i < N; i++) if (P[i] && P[i] == Q[i]) printf ("%d\n", i + 1);

	return 0;
}
### Prompt

Your challenge is to write 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>
using namespace std;

int N, K, C;
char S[200200]; int P[200200], Q[200200];

int main()
{
	scanf ("%d %d %d", &N, &K, &C);
	scanf ("%s", S);

	for (int i = 0, c = 0; i < N; i++) if (S[i] == 'o'){
		P[i] = ++c;
		i += C;
		if (c == K) break;
	}

	for (int i = N - 1, c = K; i >= 0; i--) if (S[i] == 'o'){
		Q[i] = c--;
		i -= C;
		if (c == 0) break;
	}

	for (int i = 0; i < N; i++) if (P[i] && P[i] == Q[i]) printf ("%d\n", i + 1);

	return 0;
}
```