#include <iostream>
#include <string>
#include <vector>
using namespace std;

int main(){
    int n,k,c;cin>>n>>k>>c;
    string s;cin>>s;
     int a[200000]; int b[200000];
    for(int i=0,cur=0;i<n;i++){
        if(s[i]=='o'){
            a[cur++]=i;
            i+=c;
        }
    }
    for(int i=n-1,cur=k-1;i>=0;i--){
        if(s[i]=='o'){
            b[cur--]=i;
            i-=c;
        }
    }
    for(int i=0;i<k;i++){
        if(a[i]==b[i])cout<<a[i]+1<<endl;
    }
}
### Prompt

Develop a solution in cpp to the problem described below:

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 <iostream>
#include <string>
#include <vector>
using namespace std;

int main(){
    int n,k,c;cin>>n>>k>>c;
    string s;cin>>s;
     int a[200000]; int b[200000];
    for(int i=0,cur=0;i<n;i++){
        if(s[i]=='o'){
            a[cur++]=i;
            i+=c;
        }
    }
    for(int i=n-1,cur=k-1;i>=0;i--){
        if(s[i]=='o'){
            b[cur--]=i;
            i-=c;
        }
    }
    for(int i=0;i<k;i++){
        if(a[i]==b[i])cout<<a[i]+1<<endl;
    }
}
```