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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
#include<cassert>
#include<numeric>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
const int INF = 1001001001;
const long long LINF = 1001002003004005006ll;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
typedef long long ll;
typedef pair<int,int> P;

int main(){
  int n,k,c;
  cin >> n >> k >> c;
  string s;
  cin >> s;
  vector<int> fast(n,-1);
  vector<int> slow(n,-1);
  int day = 1;
  rep(i,n){
    if(s[i]=='o'){
      fast[i]=day;
      if(day==k)break;
      day++;
      i+=c;
    }
  }
  for(int i=n-1;i>=0;i--){
    if(s[i]=='o'){
      slow[i]=day;
      day--;
      i-=c;
    }
  }
  rep(i,n){
    if(fast[i]==-1)continue;
    if(fast[i]==slow[i])cout << i+1 << endl;
  }
}