#include <bits/stdc++.h>

#define rep(i, n) for(int (i) = 0; i < (n); ++(i))

using ll = long long int;
const ll MOD = 1e9 + 7;

int main() {
  int N, K, C;
  std::string S;
  std::cin >> N >> K >> C >> S;

  std::vector<int> L(K), R(K);
  int day = 1;
  rep(i, K) {
    while(S[day - 1] == 'x') {
      ++day;
    }
    L[i] = day;
    day += C + 1;
  }

  day = N;
  rep(i, K) {
    while(S[day - 1] == 'x') {
      --day;
    }
    R[K - i - 1] = day;
    day -= C + 1;
  }

  rep(i, K) {
    if(L[i] == R[i]) {
      std::cout << L[i] << '\n';
    }
  }
}

### 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 <bits/stdc++.h>

#define rep(i, n) for(int (i) = 0; i < (n); ++(i))

using ll = long long int;
const ll MOD = 1e9 + 7;

int main() {
  int N, K, C;
  std::string S;
  std::cin >> N >> K >> C >> S;

  std::vector<int> L(K), R(K);
  int day = 1;
  rep(i, K) {
    while(S[day - 1] == 'x') {
      ++day;
    }
    L[i] = day;
    day += C + 1;
  }

  day = N;
  rep(i, K) {
    while(S[day - 1] == 'x') {
      --day;
    }
    R[K - i - 1] = day;
    day -= C + 1;
  }

  rep(i, K) {
    if(L[i] == R[i]) {
      std::cout << L[i] << '\n';
    }
  }
}

```