#include<bits/stdc++.h>
using namespace std;
string s;
int a[200005],b[200005];
int main()
{
    memset(a,0,sizeof a);
    memset(b,0,sizeof b);
    int n,k,c,jqs=1;
    cin>>n>>k>>c>>s;
    for(int i=0;i<n;i++)
    {
        if(s[i]=='o'){a[i+1]=jqs;i+=c;jqs++;}
        if(jqs>k)break;
    }
    jqs--;
    for(int i=n-1;i>=0;i--)
    {
        if(s[i]=='o'){b[i+1]=jqs;i-=c;jqs--;}
        if(jqs==0)break;
    }
    for(int i=1;i<=n;i++)
    {
        if(a[i]&&b[i]&&a[i]==b[i])cout<<i<<endl;
    }
    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;
string s;
int a[200005],b[200005];
int main()
{
    memset(a,0,sizeof a);
    memset(b,0,sizeof b);
    int n,k,c,jqs=1;
    cin>>n>>k>>c>>s;
    for(int i=0;i<n;i++)
    {
        if(s[i]=='o'){a[i+1]=jqs;i+=c;jqs++;}
        if(jqs>k)break;
    }
    jqs--;
    for(int i=n-1;i>=0;i--)
    {
        if(s[i]=='o'){b[i+1]=jqs;i-=c;jqs--;}
        if(jqs==0)break;
    }
    for(int i=1;i<=n;i++)
    {
        if(a[i]&&b[i]&&a[i]==b[i])cout<<i<<endl;
    }
    return 0;
}
```