#include <cstdio>
#include <algorithm>
#include <set>
using namespace std;
int main() {
	int N, K, C;
	scanf("%d %d %d", &N, &K, &C);
	char S[N];
	for (int i = 0; i < N; ++i) scanf(" %c", &S[i]);
	int l[K], u[K], b = N, f = K - 1;
	for (int i = N - 1; i >= 0; --i) {
		if (S[i] == 'o' && i < b) u[f] = i, b = i - C, --f;
	}
	f = 0, b = -1;
	for (int i = 0; i < N; ++i) {
		if (S[i] == 'o' && i > b) l[f] = i, b = i + C, ++f;
	}
	for (int i = 0; i < K; ++i) {
		if (l[i] == u[i]) printf("%d\n", l[i] + 1);
	}
	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 <cstdio>
#include <algorithm>
#include <set>
using namespace std;
int main() {
	int N, K, C;
	scanf("%d %d %d", &N, &K, &C);
	char S[N];
	for (int i = 0; i < N; ++i) scanf(" %c", &S[i]);
	int l[K], u[K], b = N, f = K - 1;
	for (int i = N - 1; i >= 0; --i) {
		if (S[i] == 'o' && i < b) u[f] = i, b = i - C, --f;
	}
	f = 0, b = -1;
	for (int i = 0; i < N; ++i) {
		if (S[i] == 'o' && i > b) l[f] = i, b = i + C, ++f;
	}
	for (int i = 0; i < K; ++i) {
		if (l[i] == u[i]) printf("%d\n", l[i] + 1);
	}
	return 0;
}
```