#include <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
vector<int> d1,d2;

int main()
{
	int N,K,C;
	string S;
	cin >> N >> K >> C;
	cin >> S;
	int idx = 0, cnt = 0;
	while(cnt < K)
	{
		if (S[idx] == 'o')
		{
			d1.push_back(idx);
			idx += C;
			cnt++;
		}
		idx++;
	}
	idx = N-1, cnt = 0;
	while(cnt < K)
	{
		if (S[idx] == 'o')
		{
			d2.push_back(idx);
			idx -= C;
			cnt++;
		}
		idx--;
	}
	reverse(d2.begin(),d2.end());
	rep(i,K)
	{
		if (d1[i] == d2[i]) cout << d1[i]+1 << endl;
	}
	return 0;
}
### Prompt

Develop a solution in cpp 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 <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
vector<int> d1,d2;

int main()
{
	int N,K,C;
	string S;
	cin >> N >> K >> C;
	cin >> S;
	int idx = 0, cnt = 0;
	while(cnt < K)
	{
		if (S[idx] == 'o')
		{
			d1.push_back(idx);
			idx += C;
			cnt++;
		}
		idx++;
	}
	idx = N-1, cnt = 0;
	while(cnt < K)
	{
		if (S[idx] == 'o')
		{
			d2.push_back(idx);
			idx -= C;
			cnt++;
		}
		idx--;
	}
	reverse(d2.begin(),d2.end());
	rep(i,K)
	{
		if (d1[i] == d2[i]) cout << d1[i]+1 << endl;
	}
	return 0;
}
```