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;
#include "string"
#define endl '\n'
#define PI pair<int,int>
#define ff first
#define ss second
#define int long long 
#define boost ios_base::sync_with_stdio(false);cin.tie(nullptr)

int32_t main() { 
    boost;

    int n,k,c;
    cin>>n>>k>>c;
    string s;
    cin>>s;
    int l[n+1],r[n+2];
    l[0]=r[0]=0;
    int last = -c-1;
    for(int i=1;i<=n;i++){
    	if(s[i-1]=='x')
    		l[i]=l[i-1];
    	else{
    		if(i>last+c){
    			l[i]=l[i-1]+1;
    			last=i;
    		}
    		else
    			l[i]=l[i-1];
    	}
    }
   	
   	last=n+c+1;
   	r[n+1]={0};
   	for(int i=n;i>0;i--){
   		if(s[i-1]=='x')
   			r[i]=r[i+1];
   		else {
   			if(last-i>c){
   				r[i]=r[i+1]+1;
   				last=i;
   			}
   			else
   				r[i]=r[i+1];
   		}
   	}
   	//print(l,n+1);
   	//print(r,n+1);
   	for(int i=1;i<=n;i++){
   		if(s[i-1]=='o' && l[i-1]+r[i+1]<k && l[i]+r[i]==k+1)
   			cout<<i<<endl;
   	}
}