#include <bits/stdc++.h>
using namespace std;
using int64 = long long;

int main(){
	int n, k, c;
	string s;
	cin >> n >> k >> c >> s;
	vector<int> l(k), r(k);
	int idx = 0;
	for(int i = 0; i < n; ++i){
		if(s[i] == 'x') continue;
		l[idx++] = i;
		i += c;
		if(idx == k) break;
	}
	idx = k - 1;
	for(int i = n - 1; i >= 0; --i){
		if(s[i] == 'x') continue;
		r[idx--] = i;
		i -= c;
		if(idx == -1) break;
	}
	for(int i = 0; i < k; ++i){
		if(l[i] == r[i]) cout << l[i] + 1 << endl;
	}
}
### Prompt

Generate 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;
using int64 = long long;

int main(){
	int n, k, c;
	string s;
	cin >> n >> k >> c >> s;
	vector<int> l(k), r(k);
	int idx = 0;
	for(int i = 0; i < n; ++i){
		if(s[i] == 'x') continue;
		l[idx++] = i;
		i += c;
		if(idx == k) break;
	}
	idx = k - 1;
	for(int i = n - 1; i >= 0; --i){
		if(s[i] == 'x') continue;
		r[idx--] = i;
		i -= c;
		if(idx == -1) break;
	}
	for(int i = 0; i < k; ++i){
		if(l[i] == r[i]) cout << l[i] + 1 << endl;
	}
}
```