#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;

int main(){
    int n, k, c; cin >> n >> k >> c;
    string s; cin >>  s;
    vector<int> front(n,-1), back(n,-2);

    int count = 1;
    for(int i = 0; i < n; ){
        if(s[i] == 'x'){ i++; continue; }
        front[i] = count;
        count++;
        i += c+1;
    }
    count = k;
    for(int i = n-1; i >= 0; ){
        if(s[i] == 'x'){ i--; continue; }
        back[i] = count;
        count--;
        i -= c+1;
    }
    rep(i,n) if(front[i] == back[i]) cout << i+1 << endl;
    return 0;
}
### Prompt

Please create a solution in cpp 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>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;

int main(){
    int n, k, c; cin >> n >> k >> c;
    string s; cin >>  s;
    vector<int> front(n,-1), back(n,-2);

    int count = 1;
    for(int i = 0; i < n; ){
        if(s[i] == 'x'){ i++; continue; }
        front[i] = count;
        count++;
        i += c+1;
    }
    count = k;
    for(int i = n-1; i >= 0; ){
        if(s[i] == 'x'){ i--; continue; }
        back[i] = count;
        count--;
        i -= c+1;
    }
    rep(i,n) if(front[i] == back[i]) cout << i+1 << endl;
    return 0;
}
```