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 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<stdbool.h>
#include<math.h>
#define MAX 200002
int N,K,C;
char S[MAX];
bool f[MAX];
void solve(int today,int working){
    bool can=false;
    int n = K - working,day = today;
    if(today > N || working >= K) return;
    if(today == N && K - working == 1){
        f[today] = true;
        return;
    }    
    else if(S[today] == 'x') solve(today+1,working);
    while(n > 0 && day <= N){
        if(S[day]=='o'){
            day += C+1;
            n--;
        }
        else
            day++;
    }
    if(n > 0) return;
    else{
        n = K - working,day = today + 1;
        while(n > 0 && day <= N){
            if(S[day]=='o'){
                day += C+1;
                n--;
            }
            else
                day++;
        }
        if(n > 0) f[today] = true;
        solve(today+C+1,working+1);
    }
}
int main(){
    scanf("%d %d %d",&N,&K,&C);
        scanf("%s",S+1);
    memset(f,false,sizeof(f));
    solve(1,0);
    for(int i=1;i<=N;i++){
        if(f[i] == true)
            printf("%d\n",i);
    }
    return 0;
}