#include<iostream>
using namespace std;

int main() {
    int N, K, C;
    string S;
    cin >> N >> K >> C >> S;
    int l[K], r[K], idx = 0;
    for (int i = 0; i < N; i++) {
        if (S[i] == 'x') continue;
        l[idx++] = i;
        if (idx == K) break;
        i += C;
    }
    idx = K - 1;
    for (int i = N - 1; i >= 0; i--) {
        if (S[i] == 'x') continue;
        r[idx--] = i;
        if (idx < 0) break;
        i -= C;
    }
    for (int i = 0; i < K; i++) {
        if (l[i] == r[i]) cout << l[i]+1 << endl;
    }
}
### 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<iostream>
using namespace std;

int main() {
    int N, K, C;
    string S;
    cin >> N >> K >> C >> S;
    int l[K], r[K], idx = 0;
    for (int i = 0; i < N; i++) {
        if (S[i] == 'x') continue;
        l[idx++] = i;
        if (idx == K) break;
        i += C;
    }
    idx = K - 1;
    for (int i = N - 1; i >= 0; i--) {
        if (S[i] == 'x') continue;
        r[idx--] = i;
        if (idx < 0) break;
        i -= C;
    }
    for (int i = 0; i < K; i++) {
        if (l[i] == r[i]) cout << l[i]+1 << endl;
    }
}
```