#include <bits/stdc++.h>
#define fr(x,y) for(int x=0;x<y;x++)
using namespace std;
int main() 
{
	int n,k,c;
	cin>>n>>k>>c;
	string s;
	cin>>s;
	bool* start=new bool[n],* end=new bool[n];
	int count = 0;
	fr(i,n) 
	{
		start[i]=end[i]=false;
	}
	fr(i,n)
	{
		if(s[i]=='x') continue;
		count++;
		start[i]=true;
		i=i+c;
		if(count>k) return 0;
	}
	count=0;
	for(int i=n-1;i>=0;i--) 
	{
		if(s[i]=='x') continue;
		end[i]=true;
		count++;
		i=i-c;
		if(count>k) return 0;
	}
	fr(i,n)
	{
		if (end[i]==true && start[i]==true) cout<<i+1<<endl;
	}
	return 0;
}

### Prompt

Generate 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>
#define fr(x,y) for(int x=0;x<y;x++)
using namespace std;
int main() 
{
	int n,k,c;
	cin>>n>>k>>c;
	string s;
	cin>>s;
	bool* start=new bool[n],* end=new bool[n];
	int count = 0;
	fr(i,n) 
	{
		start[i]=end[i]=false;
	}
	fr(i,n)
	{
		if(s[i]=='x') continue;
		count++;
		start[i]=true;
		i=i+c;
		if(count>k) return 0;
	}
	count=0;
	for(int i=n-1;i>=0;i--) 
	{
		if(s[i]=='x') continue;
		end[i]=true;
		count++;
		i=i-c;
		if(count>k) return 0;
	}
	fr(i,n)
	{
		if (end[i]==true && start[i]==true) cout<<i+1<<endl;
	}
	return 0;
}

```