#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;
typedef pair<ll,ll> pll;
const ll mod=1e9+7;
//const ll mod=998244353;
const ll inf=5e18;

int main() {
  ll n,k,c;
  cin >> n >> k >> c;
  string s;
  cin >> s;
  vec l(k),r(k);
  ll idk=0;
  for(ll i=0;i<n&&idk<k;i++) {
    if(s[i]=='o') {
      l[idk]=i;
      idk++;
      i+=c;
    }
  }
  idk--;
  for(ll i=n-1;i>-1&&idk>-1;i--) {
    if(s[i]=='o') {
      r[idk]=i;
      idk--;
      i-=c;
    }
  }
  for(ll i=0;i<k;i++) {
    if(l[i]==r[i]) cout << l[i]+1 << endl;
  }
}
### Prompt

Generate 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;
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;
typedef pair<ll,ll> pll;
const ll mod=1e9+7;
//const ll mod=998244353;
const ll inf=5e18;

int main() {
  ll n,k,c;
  cin >> n >> k >> c;
  string s;
  cin >> s;
  vec l(k),r(k);
  ll idk=0;
  for(ll i=0;i<n&&idk<k;i++) {
    if(s[i]=='o') {
      l[idk]=i;
      idk++;
      i+=c;
    }
  }
  idk--;
  for(ll i=n-1;i>-1&&idk>-1;i--) {
    if(s[i]=='o') {
      r[idk]=i;
      idk--;
      i-=c;
    }
  }
  for(ll i=0;i<k;i++) {
    if(l[i]==r[i]) cout << l[i]+1 << endl;
  }
}
```