#include<bits/stdc++.h>
using namespace std;
int main(){
    int n,k,c;
    cin>>n>>k>>c;
    string s;
    cin>>s;
    int l[k],r[k];
    int b=0,e=k-1;
    int ind=-1;
    for(int i=0;i<n && b<k;i++){
        if(s[i]=='o' && i>ind){
            l[b]=i;
            b++;
            ind=i+c;
        }
 
    }
 
    ind=n;
    for(int i=n-1;i>=0 && e>=0;i--){
        if(s[i]=='o' && i<ind){
            r[e]=i;
            e--;
            ind=i-c;
        }
    }
    
    for(int i=0;i<k;i++){
        if(l[i]==r[i])cout<<l[i]+1<<endl;
        
    }
 
	return 0;
}
### Prompt

In Cpp, your task is to solve 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;
int main(){
    int n,k,c;
    cin>>n>>k>>c;
    string s;
    cin>>s;
    int l[k],r[k];
    int b=0,e=k-1;
    int ind=-1;
    for(int i=0;i<n && b<k;i++){
        if(s[i]=='o' && i>ind){
            l[b]=i;
            b++;
            ind=i+c;
        }
 
    }
 
    ind=n;
    for(int i=n-1;i>=0 && e>=0;i--){
        if(s[i]=='o' && i<ind){
            r[e]=i;
            e--;
            ind=i-c;
        }
    }
    
    for(int i=0;i<k;i++){
        if(l[i]==r[i])cout<<l[i]+1<<endl;
        
    }
 
	return 0;
}
```