#include<bits/stdc++.h>
#define mod 1000000007
#define ll long long int
#define ar array
#define mp make_pair
using namespace std;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	int n,a,b;
	cin>>n>>a>>b;
	string s;
	cin>>s;
	int j = 0;vector<int>v(a),c(a);
	for(int i = 0; i<s.length(),j<a; i++)
		if(s[i]=='o')
		{
			v[j++]=i;
			i += b;
		}
	j = a-1;
	for(int i = s.length()-1; i>=0,j>=0; i--)
		if(s[i]=='o')
		{
			c[j--]= i;
			i -=b;
		}
	for(int i = 0; i<a; i++)
	{
		if(v[i]==c[i])
			cout<<v[i]+1<<endl;
	}
	return 0;
}
### Prompt

Your task is to create 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>
#define mod 1000000007
#define ll long long int
#define ar array
#define mp make_pair
using namespace std;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	int n,a,b;
	cin>>n>>a>>b;
	string s;
	cin>>s;
	int j = 0;vector<int>v(a),c(a);
	for(int i = 0; i<s.length(),j<a; i++)
		if(s[i]=='o')
		{
			v[j++]=i;
			i += b;
		}
	j = a-1;
	for(int i = s.length()-1; i>=0,j>=0; i--)
		if(s[i]=='o')
		{
			c[j--]= i;
			i -=b;
		}
	for(int i = 0; i<a; i++)
	{
		if(v[i]==c[i])
			cout<<v[i]+1<<endl;
	}
	return 0;
}
```