// editorial
#include<iostream>
#include<vector>
using namespace std;

int main(){
    int n, k, c;
    cin >> n >> k >> c;
    string s;
    cin >> s;
    vector<int> r(n, -1), l(n, -1);
    int tim = 0;
    for(int i = 0; i < n; i++){
        if(s[i] == 'x') continue;
        r[tim++] = i;
        i += c;
    }
    tim = k-1;
    for(int i = n-1; i >= 0; i--){
        if(s[i] == 'x') continue;
        l[tim--] = i;
        if(tim < 0) break;
        i -= c;
    }
    for(int i = 0; i < k; i++){
        if(r[i] == l[i])    cout << r[i]+1 << endl;
    }
    return 0;
}
### 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
// editorial
#include<iostream>
#include<vector>
using namespace std;

int main(){
    int n, k, c;
    cin >> n >> k >> c;
    string s;
    cin >> s;
    vector<int> r(n, -1), l(n, -1);
    int tim = 0;
    for(int i = 0; i < n; i++){
        if(s[i] == 'x') continue;
        r[tim++] = i;
        i += c;
    }
    tim = k-1;
    for(int i = n-1; i >= 0; i--){
        if(s[i] == 'x') continue;
        l[tim--] = i;
        if(tim < 0) break;
        i -= c;
    }
    for(int i = 0; i < k; i++){
        if(r[i] == l[i])    cout << r[i]+1 << endl;
    }
    return 0;
}
```