#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+5;
const int inf=0x3f3f3f3f;
int a[maxn],b[maxn];
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int n,k,c;
	string s;
	cin>>n>>k>>c>>s;
	int num=0; 
	for(int i=0;i<n;i++) 
	{
		if(s[i]=='o') 
		{
			a[num++]=i;
			i+=c;
		}
	}
	num=k-1;
	for(int i=n-1;i>=0;i--) 
	{
		if(num<0)
			break;
		if(s[i]=='o') 
		{
			b[num--]=i;
			i-=c;
		}
	}
	for(int i=0;i<k;i++) 
		if(a[i]==b[i])
			cout<<a[i]+1<<endl;
	return 0; 
}

### Prompt

Create a solution in Cpp for 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 maxn=1e6+5;
const int inf=0x3f3f3f3f;
int a[maxn],b[maxn];
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int n,k,c;
	string s;
	cin>>n>>k>>c>>s;
	int num=0; 
	for(int i=0;i<n;i++) 
	{
		if(s[i]=='o') 
		{
			a[num++]=i;
			i+=c;
		}
	}
	num=k-1;
	for(int i=n-1;i>=0;i--) 
	{
		if(num<0)
			break;
		if(s[i]=='o') 
		{
			b[num--]=i;
			i-=c;
		}
	}
	for(int i=0;i<k;i++) 
		if(a[i]==b[i])
			cout<<a[i]+1<<endl;
	return 0; 
}

```