#include <iostream>
#include <set>
using namespace std;
const int MAXN = 200005;
int n, k, c;
char s[MAXN];

int main()
{
    cin >> n >> k >> c;
    cin >> (s + 1);

    set<int> s1, s2;
    for (int i = 1; i <= n; i++)
    {
        if (s[i] == 'o') s1.insert(i), i += c; 
    }

    for (int j = n; j >= 1; j--)
    {
        if (s[j] == 'o') s2.insert(j), j -= c;
    }

    if (s1.size() > k || s2.size() > k) return 0;
    for (int v : s1)
    {
        if (s2.count(v)) cout << v << endl;
    }
}
### Prompt

Your challenge is to write 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 <set>
using namespace std;
const int MAXN = 200005;
int n, k, c;
char s[MAXN];

int main()
{
    cin >> n >> k >> c;
    cin >> (s + 1);

    set<int> s1, s2;
    for (int i = 1; i <= n; i++)
    {
        if (s[i] == 'o') s1.insert(i), i += c; 
    }

    for (int j = n; j >= 1; j--)
    {
        if (s[j] == 'o') s2.insert(j), j -= c;
    }

    if (s1.size() > k || s2.size() > k) return 0;
    for (int v : s1)
    {
        if (s2.count(v)) cout << v << endl;
    }
}
```