#include <cstdio>
int main() {
    int n, k, c, l[200005], r[200005], t = 0;
    char s[200005];
    scanf("%d %d %d\n%s", &n, &k, &c, s);
    int i = 0, j = n - 1;
    while (t < k) {
        while (s[i] == 'x') {
            i++;
        }
        l[t++] = i;
        i += c + 1;
    }
    t = k - 1;
    while (t >= 0) {
        while (s[j] == 'x') {
            j--;
        }
        r[t--] = j;
        j -= c + 1;
    }
    for (int i = 0; i < k; i++) {
        if (l[i] == r[i]) {
            printf("%d\n", l[i] + 1);
        }
    }
    return 0;
}
### 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 <cstdio>
int main() {
    int n, k, c, l[200005], r[200005], t = 0;
    char s[200005];
    scanf("%d %d %d\n%s", &n, &k, &c, s);
    int i = 0, j = n - 1;
    while (t < k) {
        while (s[i] == 'x') {
            i++;
        }
        l[t++] = i;
        i += c + 1;
    }
    t = k - 1;
    while (t >= 0) {
        while (s[j] == 'x') {
            j--;
        }
        r[t--] = j;
        j -= c + 1;
    }
    for (int i = 0; i < k; i++) {
        if (l[i] == r[i]) {
            printf("%d\n", l[i] + 1);
        }
    }
    return 0;
}
```