#include <stdio.h>
int n, k, c;
char s[200001];
int a[200000];
int b[200000];
int x, y;
int main() {
	scanf("%d%d%d", &n, &k, &c);
	scanf(" %s", s);
	x = -n - 1;
	for (int i = 0; i < n; i++) {
		if (x + c < i && s[i] == 'o') {
			x = i;
			a[i] = 1;
			y++;
		}
	}
	if (y > k)return 0;
	x = 2 * n;
	for (int i = n - 1; i >= 0; i--) {
		if (x - c > i && s[i] == 'o') {
			x = i;
			b[i] = 1;
		}
	}
	for (int i = 0; i < n; i++) {
		if (a[i] && b[i])printf("%d\n", i + 1);
	}
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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>
int n, k, c;
char s[200001];
int a[200000];
int b[200000];
int x, y;
int main() {
	scanf("%d%d%d", &n, &k, &c);
	scanf(" %s", s);
	x = -n - 1;
	for (int i = 0; i < n; i++) {
		if (x + c < i && s[i] == 'o') {
			x = i;
			a[i] = 1;
			y++;
		}
	}
	if (y > k)return 0;
	x = 2 * n;
	for (int i = n - 1; i >= 0; i--) {
		if (x - c > i && s[i] == 'o') {
			x = i;
			b[i] = 1;
		}
	}
	for (int i = 0; i < n; i++) {
		if (a[i] && b[i])printf("%d\n", i + 1);
	}
}
```