#include<bits/stdc++.h>
using namespace std;
int main(){
    int N,K,C;
    string str;
    cin >> N >> K >> C >> str;
    int L[K],R[K];
    int n=0;
    for(int i=0; i<K; i++){
        while(str.at(n)!='o'){
            n++;
        }
        L[i]=n;
        n+=C+1;
    }
    n=N-1;
    for(int i=K-1; i>=0; i--){
        while(str.at(n)!='o'){
            n--;
        }
        R[i]=n;
        n-=C+1;
    }
    for(int i=0; i<K; i++){
        if(L[i]==R[i]){
            cout << L[i]+1 << endl;
        }
    }
    return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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(){
    int N,K,C;
    string str;
    cin >> N >> K >> C >> str;
    int L[K],R[K];
    int n=0;
    for(int i=0; i<K; i++){
        while(str.at(n)!='o'){
            n++;
        }
        L[i]=n;
        n+=C+1;
    }
    n=N-1;
    for(int i=K-1; i>=0; i--){
        while(str.at(n)!='o'){
            n--;
        }
        R[i]=n;
        n-=C+1;
    }
    for(int i=0; i<K; i++){
        if(L[i]==R[i]){
            cout << L[i]+1 << endl;
        }
    }
    return 0;
}

```