#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)

int main()
{
ll n , k , c ;
cin >> n >> k >> c ;
string s ;
cin >> s ;
vector<ll> v(n,0);
vector<ll> vv(n,0);
ll l=-100000000;
ll cnt=1;
rep(i,n){
  if(i-l<=c)continue;
  if(s[i]=='o'){
    l=i;
    v[i]=cnt;
    cnt++;
  }
}
l=-100000000;
cnt=k;
rep(i,n){
  if(i-l<=c)continue;
  if(s[n-1-i]=='o'){
    l=i;
    vv[n-1-i]=cnt;
    cnt--;
  }
}
rep(i,n){
  if(v[i]!=0&&v[i]==vv[i])cout << i+1 << endl;
}
return 0;
}
### Prompt

Create a solution in cpp for 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;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)

int main()
{
ll n , k , c ;
cin >> n >> k >> c ;
string s ;
cin >> s ;
vector<ll> v(n,0);
vector<ll> vv(n,0);
ll l=-100000000;
ll cnt=1;
rep(i,n){
  if(i-l<=c)continue;
  if(s[i]=='o'){
    l=i;
    v[i]=cnt;
    cnt++;
  }
}
l=-100000000;
cnt=k;
rep(i,n){
  if(i-l<=c)continue;
  if(s[n-1-i]=='o'){
    l=i;
    vv[n-1-i]=cnt;
    cnt--;
  }
}
rep(i,n){
  if(v[i]!=0&&v[i]==vv[i])cout << i+1 << endl;
}
return 0;
}
```