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 For(i,n,k) for(int i=(n);i<(k);i++)
#define ALL(a)  (a).begin(),(a).end()
ll ans=0;
ll n,k,c;
vector<int> calc(string s){
    vector<int> day;
    int interval=c+1;
    For(i,0,n){
        if(s[i]=='x') {
            interval++;
        }
        else{
            if(interval<c){
                interval++;
            } 
            else{
                day.push_back(i+1);
                interval=0;
            }
        }
    }
    return day;
}
void Main(){
    cin>>n>>k>>c;
    string s;cin>>s;
    vector<int> dp=calc(s);
    if(dp.size()>k) return;
    reverse(ALL(s));
    vector<int> pd=calc(s);
    For(i,0,k){
        pd[i]=n+1-pd[i];
    }
    reverse(ALL(pd));
    For(i,0,k){
        if(pd[i]==dp[i]) cout<<dp[i]<<endl;
    }
}
int main(){
    Main();
    /*
    東方風神録は神が出てくるので当然神ゲー
    */
    return 0;
}