#include <stdio.h>

int start[200010],end[200010];
char x[200010];
int main()
{
	int a,b,c;
	scanf("%d%d%d",&a,&b,&c);
	scanf("%s",x+1);
	
	int p = 1;
	for(int i=1;i<=b;i++)
	{
		while(p<=a)
		{
			if(x[p]=='o')
			{
				start[i] = p;
				p += (c+1);
				break;
			}
			else p++;
		}
	}
	
	p = a;
	for(int i=b;i>=1;i--)
	{
		while(p>=1)
		{
			if(x[p]=='o')
			{
				end[i] = p;
				p -= (c+1);
				break;
			}
			else p--;
		}
	}
	
	for(int i=1;i<=b;i++)
	{
		if(start[i]==end[i]) printf("%d\n",start[i]);
	}
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 <stdio.h>

int start[200010],end[200010];
char x[200010];
int main()
{
	int a,b,c;
	scanf("%d%d%d",&a,&b,&c);
	scanf("%s",x+1);
	
	int p = 1;
	for(int i=1;i<=b;i++)
	{
		while(p<=a)
		{
			if(x[p]=='o')
			{
				start[i] = p;
				p += (c+1);
				break;
			}
			else p++;
		}
	}
	
	p = a;
	for(int i=b;i>=1;i--)
	{
		while(p>=1)
		{
			if(x[p]=='o')
			{
				end[i] = p;
				p -= (c+1);
				break;
			}
			else p--;
		}
	}
	
	for(int i=1;i<=b;i++)
	{
		if(start[i]==end[i]) printf("%d\n",start[i]);
	}
}
```