        #include <bits/stdc++.h>
        using namespace std;
   #define int long long
const int MAX = 510000;
const int MOD =  998244353;
const int  Inv2 = (MOD+1)/2;

     signed main(){
  int n,k,c;
  cin>>n>>k>>c;
  string s;
  cin>>s;
int t=0;
int a[n],b[n];a[0]=0;b[n-1]=0;
for(int i=0;i<n-1;i++){
   if(t>0||s[i]=='x'){a[i+1]=a[i];t--;}
   else {a[i+1]=a[i]+1;t=c;}
}
t=0;
for(int i=n-1;i>0;i--){
   if(t>0||s[i]=='x'){b[i-1]=b[i];t--;}
   else {b[i-1]=b[i]+1;t=c;}
}
for(int i=0;i<n;i++)if(a[i]+b[i]<k)cout<<i+1<<endl;
      
      
      }

        

### 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;
   #define int long long
const int MAX = 510000;
const int MOD =  998244353;
const int  Inv2 = (MOD+1)/2;

     signed main(){
  int n,k,c;
  cin>>n>>k>>c;
  string s;
  cin>>s;
int t=0;
int a[n],b[n];a[0]=0;b[n-1]=0;
for(int i=0;i<n-1;i++){
   if(t>0||s[i]=='x'){a[i+1]=a[i];t--;}
   else {a[i+1]=a[i]+1;t=c;}
}
t=0;
for(int i=n-1;i>0;i--){
   if(t>0||s[i]=='x'){b[i-1]=b[i];t--;}
   else {b[i-1]=b[i]+1;t=c;}
}
for(int i=0;i<n;i++)if(a[i]+b[i]<k)cout<<i+1<<endl;
      
      
      }

        

```