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

#define rep(i, n) for(int i=0; i<n; ++i)
#define repit(it, li) for(auto it=li.begin(); it!=li.end(); it++)
#define ll long long int

int main(){
    int n, k, c;
    cin>>n>>k>>c;
    string s;
    cin>>s;

    vector<int> w1(k), w2(k);
    for(int i=0, p=0; i<k; i++){
        while(s[p]=='x') p++;
        w1[i]=p;
        p += c+1;
    }

    for(int i=k-1, p=n-1; i>=0; i--){
        while(s[p]=='x') p--;
        w2[i]=p;
        p -= c+1;
    }

    rep(i, k) if(w1[i]==w2[i]) cout<<(w1[i]+1)<<endl;

    return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 repit(it, li) for(auto it=li.begin(); it!=li.end(); it++)
#define ll long long int

int main(){
    int n, k, c;
    cin>>n>>k>>c;
    string s;
    cin>>s;

    vector<int> w1(k), w2(k);
    for(int i=0, p=0; i<k; i++){
        while(s[p]=='x') p++;
        w1[i]=p;
        p += c+1;
    }

    for(int i=k-1, p=n-1; i>=0; i--){
        while(s[p]=='x') p--;
        w2[i]=p;
        p -= c+1;
    }

    rep(i, k) if(w1[i]==w2[i]) cout<<(w1[i]+1)<<endl;

    return 0;
}

```