#include <bits/stdc++.h>

using namespace std;

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n,k,c;
    string s;
    cin >> n >> k >> c >> s;
    vector<int> one,two;
    for(int i=0;i<n;i++)
    {
        if((int)one.size()<k&&s[i]=='o')
        {
            one.push_back(i);
            i+=c;
        }
    }
    for(int i=n-1;i>=0;i--)
    {
        if((int)two.size()<k&&s[i]=='o')
        {
            two.push_back(i);
            i-=c;
        }
    }
    reverse(two.begin(),two.end());
    for(int i=0;i<k;i++) if(one[i]==two[i]) cout << one[i]+1 << "\n";
    return 0;
}

### Prompt

Please formulate 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;

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n,k,c;
    string s;
    cin >> n >> k >> c >> s;
    vector<int> one,two;
    for(int i=0;i<n;i++)
    {
        if((int)one.size()<k&&s[i]=='o')
        {
            one.push_back(i);
            i+=c;
        }
    }
    for(int i=n-1;i>=0;i--)
    {
        if((int)two.size()<k&&s[i]=='o')
        {
            two.push_back(i);
            i-=c;
        }
    }
    reverse(two.begin(),two.end());
    for(int i=0;i<k;i++) if(one[i]==two[i]) cout << one[i]+1 << "\n";
    return 0;
}

```