#include <bits/stdc++.h>
using namespace std;

template<typename T>
void chmax(T &a,T b){
  if(a<b)a=b;
}

signed main(){
  int n,k,c;cin>>n>>k>>c;
  string s;cin>>s;
  vector<int> v(n,0);
  int cnt=0;
  for(int i=n-1;i>=0;i--){
    if(s[i]=='o'){
      cnt++;
      v[i]=cnt;
      i-=c;
    }
  }
  for(int i=n-1;i>0;i--)chmax(v[i-1],v[i]);
  int now=0;
  for(int i=0;i<n;i++){
    if(i<n-1){
      if(s[i]=='o'){
        if(v[i+1]+now<k)cout<<i+1<<" ";
        now++;
        i+=c;
      }
    }
    else{
      if(s[i]=='o'&&now<k)cout<<i+1;
    }
  }
  cout<<endl;
}

### Prompt

In Cpp, your task is to solve 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;

template<typename T>
void chmax(T &a,T b){
  if(a<b)a=b;
}

signed main(){
  int n,k,c;cin>>n>>k>>c;
  string s;cin>>s;
  vector<int> v(n,0);
  int cnt=0;
  for(int i=n-1;i>=0;i--){
    if(s[i]=='o'){
      cnt++;
      v[i]=cnt;
      i-=c;
    }
  }
  for(int i=n-1;i>0;i--)chmax(v[i-1],v[i]);
  int now=0;
  for(int i=0;i<n;i++){
    if(i<n-1){
      if(s[i]=='o'){
        if(v[i+1]+now<k)cout<<i+1<<" ";
        now++;
        i+=c;
      }
    }
    else{
      if(s[i]=='o'&&now<k)cout<<i+1;
    }
  }
  cout<<endl;
}

```