#include <algorithm>
#include <string.h>
#include <iostream>
#include <stdio.h>

using namespace std;

const int N = 2e5 + 10;
const int INF = 1e9;
int n, k, c;
char s[N];
int l[N], r[N];

int main() {
	cin >> n >> k >> c;
	cin >> s + 1;
	int pre = -INF, j = 0;
	for (int i = 1; i <= n; i++) if (s[i] == 'o' && i - pre - 1 >= c) pre = i, l[++j] = i;
	pre = INF, j = k + 1;
	for (int i = n; i >= 1; i--) if (s[i] == 'o' && pre - i - 1 >= c && j >= 1) pre = i, r[--j] = i;

	for (int i = 1; i <= k; i++) if (l[i] == r[i]) cout << l[i] << endl;

	return 0;
}
### Prompt

Create a solution in cpp for 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 <algorithm>
#include <string.h>
#include <iostream>
#include <stdio.h>

using namespace std;

const int N = 2e5 + 10;
const int INF = 1e9;
int n, k, c;
char s[N];
int l[N], r[N];

int main() {
	cin >> n >> k >> c;
	cin >> s + 1;
	int pre = -INF, j = 0;
	for (int i = 1; i <= n; i++) if (s[i] == 'o' && i - pre - 1 >= c) pre = i, l[++j] = i;
	pre = INF, j = k + 1;
	for (int i = n; i >= 1; i--) if (s[i] == 'o' && pre - i - 1 >= c && j >= 1) pre = i, r[--j] = i;

	for (int i = 1; i <= k; i++) if (l[i] == r[i]) cout << l[i] << endl;

	return 0;
}
```