#include <iostream>
#include <algorithm>
#include <map>
#include <vector>
#include <cmath>
#define ll long long
using namespace std;

int main(){
	int n, k, c, mx=0, l=-2020200;
	string s;
	cin >> n >> k >> c >> s;
	vector<int> v(n,0), v2(n,0);

	for (int i=0; i<n; i++){
		if (s[i]=='o' && l+c<i){
			mx++;
			v[i]=1;
			l=i;
		}
	}

	l = 2020200;
	for (int i=n-1; i>=0; i--){
		if (s[i]=='o' && l-c>i){
			v2[i]=1;
			l=i;
		}
	}

	if (mx<=k)
		for (int i=0; i<n; i++){
			if (v[i] && v2[i])
				cout << i+1 << endl;
		}
    return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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 <iostream>
#include <algorithm>
#include <map>
#include <vector>
#include <cmath>
#define ll long long
using namespace std;

int main(){
	int n, k, c, mx=0, l=-2020200;
	string s;
	cin >> n >> k >> c >> s;
	vector<int> v(n,0), v2(n,0);

	for (int i=0; i<n; i++){
		if (s[i]=='o' && l+c<i){
			mx++;
			v[i]=1;
			l=i;
		}
	}

	l = 2020200;
	for (int i=n-1; i>=0; i--){
		if (s[i]=='o' && l-c>i){
			v2[i]=1;
			l=i;
		}
	}

	if (mx<=k)
		for (int i=0; i<n; i++){
			if (v[i] && v2[i])
				cout << i+1 << endl;
		}
    return 0;
}
```