#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i=0; i<n; i++)
#define pb push_back
#define int long long

int N, K, C;
string S;
int L[200100], R[200100];

signed main() {
    cin.tie(0); ios::sync_with_stdio(false);
    cin >> N >> K >> C >> S;
    int now = 0;
    rep(i, K) {
        while (S[now]=='x') now++;
        L[i] = now;
        now += C+1;
    }
    now = N-1;
    for (int i=K-1; i>=0; i--) {
        while (S[now]=='x') now--;
        R[i] = now;
        now -= C+1;
    }
    rep(i, K) if (L[i]==R[i]) cout << L[i]+1 << 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 <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i=0; i<n; i++)
#define pb push_back
#define int long long

int N, K, C;
string S;
int L[200100], R[200100];

signed main() {
    cin.tie(0); ios::sync_with_stdio(false);
    cin >> N >> K >> C >> S;
    int now = 0;
    rep(i, K) {
        while (S[now]=='x') now++;
        L[i] = now;
        now += C+1;
    }
    now = N-1;
    for (int i=K-1; i>=0; i--) {
        while (S[now]=='x') now--;
        R[i] = now;
        now -= C+1;
    }
    rep(i, K) if (L[i]==R[i]) cout << L[i]+1 << endl;
}
```