#include <iostream>
#include <string>
#define MAXN 200000
using namespace std;
int main(){
	int N, K, C;
	string s;
	int bk[MAXN];
	cin >> N >> K >> C >> s;
	int last = s.size()+C;
	for(int i=s.size()-1; i>=0; i--)
		if(s[i] == 'o' && last > i+C)
			bk[last=i] = K--;
	//for(int i=0; i<s.size(); i++)
		//cout << bk[i] << " ";
	//cout << endl;
	K = 1;
	last = -C-1;
	for(int i=0; i<s.size(); i++)
		if(s[i] == 'o' && last < i-C)
			if(K++ == bk[last=i])
				cout << i+1 << endl;
}






### Prompt

Create a solution in Cpp for 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 <iostream>
#include <string>
#define MAXN 200000
using namespace std;
int main(){
	int N, K, C;
	string s;
	int bk[MAXN];
	cin >> N >> K >> C >> s;
	int last = s.size()+C;
	for(int i=s.size()-1; i>=0; i--)
		if(s[i] == 'o' && last > i+C)
			bk[last=i] = K--;
	//for(int i=0; i<s.size(); i++)
		//cout << bk[i] << " ";
	//cout << endl;
	K = 1;
	last = -C-1;
	for(int i=0; i<s.size(); i++)
		if(s[i] == 'o' && last < i-C)
			if(K++ == bk[last=i])
				cout << i+1 << endl;
}






```