#include <bits/stdc++.h>

const int maxn = 2e5 + 5;
char str[maxn];
int a[maxn], b[maxn];
int n, k, c, cnt = 0, tot = 0;
int main(){
	scanf("%d%d%d",&n,&k,&c);
	scanf("%s", str+1);
	for(int i=1;i<=n;i++){
		if(str[i] == 'o' && cnt == 0){
			a[++tot] = i; cnt = c+1;
		}
		if(tot == k) break;
		if(cnt) cnt --;
	}
	tot = k, cnt = 0;
	for(int i=n;i;i--){
		if(str[i] == 'o' && cnt == 0){
			b[tot--] = i; cnt = c+1;
		}
		if(tot == 0) break;
		if(cnt) cnt --;
	}
	for(int i=1;i<=k;i++){
		if(a[i] == b[i]) printf("%d\n", a[i]);
	}
	return 0;
}
### Prompt

Your challenge is to write 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>

const int maxn = 2e5 + 5;
char str[maxn];
int a[maxn], b[maxn];
int n, k, c, cnt = 0, tot = 0;
int main(){
	scanf("%d%d%d",&n,&k,&c);
	scanf("%s", str+1);
	for(int i=1;i<=n;i++){
		if(str[i] == 'o' && cnt == 0){
			a[++tot] = i; cnt = c+1;
		}
		if(tot == k) break;
		if(cnt) cnt --;
	}
	tot = k, cnt = 0;
	for(int i=n;i;i--){
		if(str[i] == 'o' && cnt == 0){
			b[tot--] = i; cnt = c+1;
		}
		if(tot == 0) break;
		if(cnt) cnt --;
	}
	for(int i=1;i<=k;i++){
		if(a[i] == b[i]) printf("%d\n", a[i]);
	}
	return 0;
}
```