#include<bits/stdc++.h>
using namespace std;
const int N=200010;
int n,k,c;
char s[N];
int f[N],b[N];
int main()
{
	scanf("%d%d%d%s",&n,&k,&c,s+1);
	f[0]=b[0]=1;
	for(int i=1;i<=n;i+=c+1){
		while(s[i]=='x') ++i;
		f[f[0]++]=i;
	}
	for(int i=n;i>=1;i-=c+1){
		while(s[i]=='x') --i;
		b[b[0]++]=i;
	}
//	for(int i=1;i<f[0];++i) printf("f[%d]=%d ",i,f[i]);
//	printf("\n");
//	for(int i=1;i<b[0];++i) printf("b[%d]=%d ",i,b[i]);
//	printf("\n");
	for(int i=1;i<=k;++i){
		if(f[i]==b[k+1-i]) printf("%d\n",f[i]);
	}
}
### Prompt

Your task is to create 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=200010;
int n,k,c;
char s[N];
int f[N],b[N];
int main()
{
	scanf("%d%d%d%s",&n,&k,&c,s+1);
	f[0]=b[0]=1;
	for(int i=1;i<=n;i+=c+1){
		while(s[i]=='x') ++i;
		f[f[0]++]=i;
	}
	for(int i=n;i>=1;i-=c+1){
		while(s[i]=='x') --i;
		b[b[0]++]=i;
	}
//	for(int i=1;i<f[0];++i) printf("f[%d]=%d ",i,f[i]);
//	printf("\n");
//	for(int i=1;i<b[0];++i) printf("b[%d]=%d ",i,b[i]);
//	printf("\n");
	for(int i=1;i<=k;++i){
		if(f[i]==b[k+1-i]) printf("%d\n",f[i]);
	}
}
```