// practice with Dukkha
#include <algorithm>
#include <iostream>

using namespace std;

const int N = 200000;

int pp[N], qq[N];

int main() {
	int n, k, c; cin >> n >> k >> c;
	string s; cin >> s;
	for (int h = 0, i = 0, o = i - c - 1; i < n; i++)
		if (s[i] == 'o' && i - o > c) {
			pp[h] = o = i;
			if (++h == k)
				break;
		}
	for (int h = 0, i = n - 1, o = i + c + 1; i >= 0; i--)
		if (s[i] == 'o' && o - i > c) {
			qq[k - 1 - h] = o = i;
			if (++h == k)
				break;
		}
	for (int h = 0; h < k; h++)
		if (pp[h] == qq[h])
			cout << pp[h] + 1 << '\n';
	return 0;
}

### Prompt

Your challenge is to write 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
// practice with Dukkha
#include <algorithm>
#include <iostream>

using namespace std;

const int N = 200000;

int pp[N], qq[N];

int main() {
	int n, k, c; cin >> n >> k >> c;
	string s; cin >> s;
	for (int h = 0, i = 0, o = i - c - 1; i < n; i++)
		if (s[i] == 'o' && i - o > c) {
			pp[h] = o = i;
			if (++h == k)
				break;
		}
	for (int h = 0, i = n - 1, o = i + c + 1; i >= 0; i--)
		if (s[i] == 'o' && o - i > c) {
			qq[k - 1 - h] = o = i;
			if (++h == k)
				break;
		}
	for (int h = 0; h < k; h++)
		if (pp[h] == qq[h])
			cout << pp[h] + 1 << '\n';
	return 0;
}

```