#include <bits/stdc++.h>
using namespace std;
const int N=2e5+5;
int n,k,c,len,tot;
int l[N],r[N];
char str[N];
int main(){
	scanf("%d%d%d",&n,&k,&c);
	scanf("%s",str+1);
	len=strlen(str+1);
	int i=1;
	while (i<=len)
	{
		if (tot==k) break;
		if (str[i]=='o') l[++tot]=i,i+=c+1;
		else i++;
	}
	i=len;
	tot=0;
	while (i>=1)
	{
		if (tot==k) break;
		if (str[i]=='o') r[++tot]=i,i-=c+1;
		else i--;
	}
	reverse(r+1,r+tot+1);
	for (register int i=1; i<=tot; ++i) if (l[i]==r[i]) printf("%d\n",l[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>
using namespace std;
const int N=2e5+5;
int n,k,c,len,tot;
int l[N],r[N];
char str[N];
int main(){
	scanf("%d%d%d",&n,&k,&c);
	scanf("%s",str+1);
	len=strlen(str+1);
	int i=1;
	while (i<=len)
	{
		if (tot==k) break;
		if (str[i]=='o') l[++tot]=i,i+=c+1;
		else i++;
	}
	i=len;
	tot=0;
	while (i>=1)
	{
		if (tot==k) break;
		if (str[i]=='o') r[++tot]=i,i-=c+1;
		else i--;
	}
	reverse(r+1,r+tot+1);
	for (register int i=1; i<=tot; ++i) if (l[i]==r[i]) printf("%d\n",l[i]);
return 0;
}
```