// abc161_e2.cc
#include <bits/stdc++.h>
using namespace std;

int main() {

	int n, k, c;
	cin >> n >> k >> c;
	string s;
	cin >> s;

	vector<int> ans1, ans2;

	for (int i = 0; i < n; i++) {
		if (s[i] == 'o') {
			ans1.push_back(i);
			i += c;
		}
	}

	for (int i = n - 1; i >= 0; i--) {
		if (s[i] == 'o') {
			ans2.push_back(i);
			i -= c;
		}
	}

	if (ans1.size() == k) {
		for (auto n : ans1) {
			if (binary_search(ans2.rbegin(), ans2.rend(), n)) {
				cout << n + 1 << endl;
			}
		}
	}

	return 0;
}
### 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
// abc161_e2.cc
#include <bits/stdc++.h>
using namespace std;

int main() {

	int n, k, c;
	cin >> n >> k >> c;
	string s;
	cin >> s;

	vector<int> ans1, ans2;

	for (int i = 0; i < n; i++) {
		if (s[i] == 'o') {
			ans1.push_back(i);
			i += c;
		}
	}

	for (int i = n - 1; i >= 0; i--) {
		if (s[i] == 'o') {
			ans2.push_back(i);
			i -= c;
		}
	}

	if (ans1.size() == k) {
		for (auto n : ans1) {
			if (binary_search(ans2.rbegin(), ans2.rend(), n)) {
				cout << n + 1 << endl;
			}
		}
	}

	return 0;
}
```