#include <iostream>
#include <string>
using namespace std;

int main()
{
    int n, k, c;
    cin >> n >> k >> c;
    c++;
    string s;
    cin >> s;
    int f[200005], l[200005];
    int j = 0;
    for(int i = 0; i < k; i++){
        while(s[j] == 'x') j++;
        f[i] = j;
        j += c;
    }
    j = n - 1;
    for(int i = k - 1; i >= 0; i--){
        while(s[j] == 'x') j--;
        l[i] = j;
        j -= c;
    }
    for(int i = 0; i < k; i++){
        if(f[i] == l[i]) cout << f[i] + 1 << endl;
    }
}

### 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
#include <iostream>
#include <string>
using namespace std;

int main()
{
    int n, k, c;
    cin >> n >> k >> c;
    c++;
    string s;
    cin >> s;
    int f[200005], l[200005];
    int j = 0;
    for(int i = 0; i < k; i++){
        while(s[j] == 'x') j++;
        f[i] = j;
        j += c;
    }
    j = n - 1;
    for(int i = k - 1; i >= 0; i--){
        while(s[j] == 'x') j--;
        l[i] = j;
        j -= c;
    }
    for(int i = 0; i < k; i++){
        if(f[i] == l[i]) cout << f[i] + 1 << endl;
    }
}

```