#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 2e5 + 10;
const int INF = 0x3f3f3f3f;
int n, k, c;
char s[N];
int L[N], R[N];
int main() {
	scanf("%d%d%d", &n, &k, &c);
	scanf("%s", s + 1);
	int cur = 1, d = 0;
	while (cur <= n) {
		if (s[cur] == 'o') {
			L[++d] = cur;
			cur += c;
		}
		cur++;
	}
	cur = n, d = 0;
	while (cur >= 1) {
		if (s[cur] == 'o') {
			R[++d] = cur;
			cur -= c;
		}
		cur--;
	}
	if (L[k + 1])puts("");
	else {
		reverse(R + 1, R + d + 1);
		for (int i = 1; i <= d; i++) {
			if (L[i] == R[i])printf("%d\n", L[i]);
		}
	}
	return 0;
}
### Prompt

Please formulate 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<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 2e5 + 10;
const int INF = 0x3f3f3f3f;
int n, k, c;
char s[N];
int L[N], R[N];
int main() {
	scanf("%d%d%d", &n, &k, &c);
	scanf("%s", s + 1);
	int cur = 1, d = 0;
	while (cur <= n) {
		if (s[cur] == 'o') {
			L[++d] = cur;
			cur += c;
		}
		cur++;
	}
	cur = n, d = 0;
	while (cur >= 1) {
		if (s[cur] == 'o') {
			R[++d] = cur;
			cur -= c;
		}
		cur--;
	}
	if (L[k + 1])puts("");
	else {
		reverse(R + 1, R + d + 1);
		for (int i = 1; i <= d; i++) {
			if (L[i] == R[i])printf("%d\n", L[i]);
		}
	}
	return 0;
}
```