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

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N,K,C; string S; cin >> N >> K >> C >> S;
    vector<int> dp(N+1,0);
    for (int i=N-1;i>=0;--i){
        if (S[i]=='o') dp[i]=dp[min(N,i+C+1)]+1;
        else dp[i]=dp[i+1];
    }
    int now=0;
    for (int i=0;i<N;++i){
        if (S[i]=='o'){
            if (now+dp[i+1]<K){
                cout << i+1 << '\n';
            }
            i+=C; ++now;
        }
    }
}
### 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(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N,K,C; string S; cin >> N >> K >> C >> S;
    vector<int> dp(N+1,0);
    for (int i=N-1;i>=0;--i){
        if (S[i]=='o') dp[i]=dp[min(N,i+C+1)]+1;
        else dp[i]=dp[i+1];
    }
    int now=0;
    for (int i=0;i<N;++i){
        if (S[i]=='o'){
            if (now+dp[i+1]<K){
                cout << i+1 << '\n';
            }
            i+=C; ++now;
        }
    }
}
```