#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
constexpr char sp = ' ';

int N, K, C, L[222222], R[222222];
string S;

int main() {
	cin >> N >> K >> C >> S;
	int lim = 0;
	for (int i = 1; i <= N; i++) {
		L[i] = L[i - 1];
		if (S[i - 1] == 'o'&&lim < i) {
			L[i]++;
			lim = i + C;
		}
	}
	lim = 0xE869120;
	for (int i = N; i >= 1; i--) {
		R[i] = R[i + 1];
		if (S[i - 1] == 'o'&&i < lim) {
			R[i]++;
			lim = i - C;
		}
	}
	for (int i = 1; i <= N; i++) {
		if (L[i - 1] + R[i + 1] < K) {
			cout << i << endl;
		}
	}
}
### Prompt

Please create a solution in Cpp 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;
typedef long long ll;
constexpr char sp = ' ';

int N, K, C, L[222222], R[222222];
string S;

int main() {
	cin >> N >> K >> C >> S;
	int lim = 0;
	for (int i = 1; i <= N; i++) {
		L[i] = L[i - 1];
		if (S[i - 1] == 'o'&&lim < i) {
			L[i]++;
			lim = i + C;
		}
	}
	lim = 0xE869120;
	for (int i = N; i >= 1; i--) {
		R[i] = R[i + 1];
		if (S[i - 1] == 'o'&&i < lim) {
			R[i]++;
			lim = i - C;
		}
	}
	for (int i = 1; i <= N; i++) {
		if (L[i - 1] + R[i + 1] < K) {
			cout << i << endl;
		}
	}
}
```