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

int main() {
  int N, K, C; string S; cin >> N >> K >> C >> S;
  bool f[N], l[N]; for (int i = 0; i < N; i++) f[i] = l[i] = false;
  int d = 0, m = 0;
  while (d < N) {
    if (S[d] == 'x') { d++; continue; }
    f[d] = true; m++; d += C + 1;
  }
  if (m > K) cout << endl;
  else {
    d = N - 1;
    while (d >= 0) {
      if (S[d] == 'x') { d--; continue; }
      l[d] = true; d -= C + 1;
    }
    for (int i = 0; i < N; i++)
      if (f[i] && l[i])
        cout << i + 1 << endl;
  }
}

### Prompt

Please formulate 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; string S; cin >> N >> K >> C >> S;
  bool f[N], l[N]; for (int i = 0; i < N; i++) f[i] = l[i] = false;
  int d = 0, m = 0;
  while (d < N) {
    if (S[d] == 'x') { d++; continue; }
    f[d] = true; m++; d += C + 1;
  }
  if (m > K) cout << endl;
  else {
    d = N - 1;
    while (d >= 0) {
      if (S[d] == 'x') { d--; continue; }
      l[d] = true; d -= C + 1;
    }
    for (int i = 0; i < N; i++)
      if (f[i] && l[i])
        cout << i + 1 << endl;
  }
}

```