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

int d[444444], e[222222];
int main() {
    int n, k, c;
    string s;
    cin >> n >> k >> c >> s;
    for(int i = n; i--;)
        d[i] = s[i] == 'o' ? d[i + c + 1] + 1 : d[i + 1];
    for(int i = 0; i < n; i++) {
        if(s[i] == 'o')
            e[i] = 1, i += c;
    }
    for(int i = 0; i < n; i++)
        if(e[i] && d[0] + d[i + 1] - d[i] < k)
            cout << i + 1 << '\n';
    cout << flush;
    return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 d[444444], e[222222];
int main() {
    int n, k, c;
    string s;
    cin >> n >> k >> c >> s;
    for(int i = n; i--;)
        d[i] = s[i] == 'o' ? d[i + c + 1] + 1 : d[i + 1];
    for(int i = 0; i < n; i++) {
        if(s[i] == 'o')
            e[i] = 1, i += c;
    }
    for(int i = 0; i < n; i++)
        if(e[i] && d[0] + d[i + 1] - d[i] < k)
            cout << i + 1 << '\n';
    cout << flush;
    return 0;
}

```