#include<bits/stdc++.h>
using namespace std;
const int N=2e5+9;
int n,k,c;int tot,tot1,ans,a[N],b[N],g[N]; int rest;
char s[N];
int main(){
	cin>>n>>k>>c>>s+1;
	rest=0;
	for(int i=1;i<=n;i++){
		if(s[i]=='o'&&rest<=0) a[++tot]=i,rest=c;
		else rest--;
	}
	rest=0;
	for(int i=n;i>=1;i--){
		if(s[i]=='o'&&rest<=0) b[++tot1]=i,rest=c;
		else rest--;
	}
	for(int i=1;i<=tot1/2;i++) swap(b[i],b[tot1-i+1]);
	//for(int i=1;i<=tot;i++) cout<<a[i]<<" ";
	for(int i=1;i<=min(tot,tot1);i++)
		if(a[i]==b[i]) g[++ans]=a[i];
	if(ans<=k) for(int i=1;i<=ans;i++) printf("%lld\n",g[i]);
	return 0;
}
### Prompt

Please formulate 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+9;
int n,k,c;int tot,tot1,ans,a[N],b[N],g[N]; int rest;
char s[N];
int main(){
	cin>>n>>k>>c>>s+1;
	rest=0;
	for(int i=1;i<=n;i++){
		if(s[i]=='o'&&rest<=0) a[++tot]=i,rest=c;
		else rest--;
	}
	rest=0;
	for(int i=n;i>=1;i--){
		if(s[i]=='o'&&rest<=0) b[++tot1]=i,rest=c;
		else rest--;
	}
	for(int i=1;i<=tot1/2;i++) swap(b[i],b[tot1-i+1]);
	//for(int i=1;i<=tot;i++) cout<<a[i]<<" ";
	for(int i=1;i<=min(tot,tot1);i++)
		if(a[i]==b[i]) g[++ans]=a[i];
	if(ans<=k) for(int i=1;i<=ans;i++) printf("%lld\n",g[i]);
	return 0;
}
```