#include<bits/stdc++.h>
using namespace std;

using LL = long long;
constexpr int MAXN = 2e5 + 5;
int N, K, C, a[MAXN], b[MAXN];
string s;
int main(){
    cin >> N >> K >> C;
    cin >> s;
    int u = 0, v = K - 1;
    for (int i = 0; i < N; ++ i){
            if (s[i] == 'o') {a[u++] = i; i += C;}
            if (u == K) break;
    }
    for (int i = N - 1; i >= 0; -- i){
        if (s[i] == 'o') {b[v--] = i; i -= C;}
        if (v == -1) break;
    }
    for (int i = 0; i < K; ++ i){
        if (a[i] == b[i]) cout << a[i] + 1<< endl;
    }
    return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;

using LL = long long;
constexpr int MAXN = 2e5 + 5;
int N, K, C, a[MAXN], b[MAXN];
string s;
int main(){
    cin >> N >> K >> C;
    cin >> s;
    int u = 0, v = K - 1;
    for (int i = 0; i < N; ++ i){
            if (s[i] == 'o') {a[u++] = i; i += C;}
            if (u == K) break;
    }
    for (int i = N - 1; i >= 0; -- i){
        if (s[i] == 'o') {b[v--] = i; i -= C;}
        if (v == -1) break;
    }
    for (int i = 0; i < K; ++ i){
        if (a[i] == b[i]) cout << a[i] + 1<< endl;
    }
    return 0;
}

```