#include<bits/stdc++.h>
using namespace std;
#define int long long int


int32_t main()
{
     int n,k,c; cin>>n>>k>>c;
     string s; cin>>s;
     int f[k],b[k];
     int i=0,j=0;
     while(i<n && j<k)
     {
         if(s[i]=='o'){
            f[j++] = i;
            i = i+c+1;
         }
         else i++;
     }
     i = n-1,j=k-1;
     while(i>=0 && j>=0)
     {
         if(s[i] == 'o'){
            b[j--]=i;
            i-=(c+1);
         }
         else i--;
     }
     for(int i=0;i<k;i++){
        if(f[i] == b[i]){
            cout<<f[i]+1<<endl;
        }
     }
}

### Prompt

Your challenge is to write 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;
#define int long long int


int32_t main()
{
     int n,k,c; cin>>n>>k>>c;
     string s; cin>>s;
     int f[k],b[k];
     int i=0,j=0;
     while(i<n && j<k)
     {
         if(s[i]=='o'){
            f[j++] = i;
            i = i+c+1;
         }
         else i++;
     }
     i = n-1,j=k-1;
     while(i>=0 && j>=0)
     {
         if(s[i] == 'o'){
            b[j--]=i;
            i-=(c+1);
         }
         else i--;
     }
     for(int i=0;i<k;i++){
        if(f[i] == b[i]){
            cout<<f[i]+1<<endl;
        }
     }
}

```