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 <cstdio>
#include <vector>

const int MAXN=2e5+10;

using namespace std;


int N,K,C;

char buf[MAXN];

int cnt[MAXN];

int main(){

    scanf("%d%d%d\n",&N,&K,&C);
    scanf("%s",buf+1);

    cnt[N]=(buf[N]=='o');
    for(int i=N-1;i>=0;i--){
        int pre=i+C+1;
        if (pre>N){
            if (buf[i]=='o') cnt[i]=1; else cnt[i]=cnt[i+1];
        }else{
            if (buf[i]=='o') cnt[i]=1+cnt[pre]; else cnt[i]=cnt[i+1];
        }
    }
    
    if (cnt[1]==K){
        for(int i=1;i<=N;i+=(C+1)){
            while(buf[i]!='o') i++;
            int j=i+1;
            bool antoher_exists=false;
            while(cnt[j]==cnt[i]&&j<=N){
                if (buf[j]=='o'){
                    antoher_exists=true;
                    break;
                }
                j++;
            }
            if (!antoher_exists) printf("%d\n",i);
        }
    }

    return 0;
}