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

const int N = 2e5 + 10;
char s[N];
vector<int> a, b;
int main()
{
	int n, k, c;
	scanf("%d%d%d", &n, &k, &c);
	scanf("%s", s + 1);
	int len = strlen(s + 1);
	
	
	for(int i = 1; i <= len; i++) {
		if(s[i] == 'o') {
			a.push_back(i);
			i += c;
		}
		if(a.size() == k) break;
	}
	for(int i = n; i >= 1; i--) {
		if(s[i] == 'o') {
			b.push_back(i);
			i -= c;
		}
		if(b.size() == k) break;
	}
	reverse(b.begin(), b.end());
	
	int la = a.size(), lb = b.size();
	if(la == lb) {
		for(int i = 0; i < la; i++) 
			if(a[i] == b[i]) printf("%d\n", a[i]);
	}
} 
### Prompt

Please create a solution in cpp 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;

const int N = 2e5 + 10;
char s[N];
vector<int> a, b;
int main()
{
	int n, k, c;
	scanf("%d%d%d", &n, &k, &c);
	scanf("%s", s + 1);
	int len = strlen(s + 1);
	
	
	for(int i = 1; i <= len; i++) {
		if(s[i] == 'o') {
			a.push_back(i);
			i += c;
		}
		if(a.size() == k) break;
	}
	for(int i = n; i >= 1; i--) {
		if(s[i] == 'o') {
			b.push_back(i);
			i -= c;
		}
		if(b.size() == k) break;
	}
	reverse(b.begin(), b.end());
	
	int la = a.size(), lb = b.size();
	if(la == lb) {
		for(int i = 0; i < la; i++) 
			if(a[i] == b[i]) printf("%d\n", a[i]);
	}
} 
```