#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N=2e5+5;
char s[N];
vector<int> L,R;
int main() {
    //freopen("in.txt","r",stdin);
    int n,k,c;
    scanf("%d%d%d%s",&n,&k,&c,s+1);
    for(int i=1;i<=n;) {
        if(s[i]=='o') L.push_back(i),i+=c;
        i++;
        if(L.size()==k) break;
    }
    for(int i=n;i>=1;) {
        if(s[i]=='o') R.push_back(i),i-=c;
        i--;
        if(R.size()==k) break;
    }
    for(int i=0;i<k;i++) {
        if(L[i]==R[k-i-1]) printf("%d\n",L[i]);
    }
    return 0;
}
### Prompt

Your task is to create 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>
#define ll long long
using namespace std;
const int N=2e5+5;
char s[N];
vector<int> L,R;
int main() {
    //freopen("in.txt","r",stdin);
    int n,k,c;
    scanf("%d%d%d%s",&n,&k,&c,s+1);
    for(int i=1;i<=n;) {
        if(s[i]=='o') L.push_back(i),i+=c;
        i++;
        if(L.size()==k) break;
    }
    for(int i=n;i>=1;) {
        if(s[i]=='o') R.push_back(i),i-=c;
        i--;
        if(R.size()==k) break;
    }
    for(int i=0;i<k;i++) {
        if(L[i]==R[k-i-1]) printf("%d\n",L[i]);
    }
    return 0;
}
```