Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

11 3 2
ooxxxoxxxoo

Sample Output 1

6

Takahashi is going to work on 3 days out of the 11 days. After working for a day, he will refrain from working on the subsequent 2 days.

There are four possible choices for his workdays: Day 1,6,10, Day 1,6,11, Day 2,6,10, and Day 2,6,11.

Thus, he is bound to work on Day 6.

Sample Input 2

5 2 3
ooxoo

Sample Output 2

1
5

There is only one possible choice for his workdays: Day 1,5.

Sample Input 3

5 1 0
ooooo

Sample Output 3

There may be no days on which he is bound to work.

Sample Input 4

16 4 3
ooxxoxoxxxoxoxxo

Sample Output 4

11
16

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define fi first
#define se second
#define pb push_back
#define forn(i,n) for(int i=0;i<(n);i++)
#define for1(i,n) for(int i=1;i<=n;i++)
#define forr(i,n) for(int i=n;i>=0;i--)
#define all(x) x.begin(), x.end()
const int MAXN = 1e5 +5;

void fio(){
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}

int main(){
  fio();
  int n,k,c; cin>>n>>k>>c;
  string s; cin>>s;
  unordered_map<int,int> m;
  for(int i = n-1;i>=0;i--){
    if(s[i]=='o'){
      m[i] = k--;
      i-=c;
    }
  }
  int x(0);
  for(int i=0;i<n;i++){
    if(s[i]=='o'){
      if(m[i]==++x) cout<<i+1<<endl;
      i+=c;
    }
  }
  
  return 0;
}