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: Wrong Answer)

#include<stdio.h>
#define N_MAX 200000

int hantei(long a,long b[],long c){
  for(long y=0;y<c;y++){
    if(b[y]==a)return 0;
  };
  return 1;
};

int main(){
  int magic=1;
  long N,K,C,m,p=0;
  char kago;
  char S[N_MAX]={};
  long hako[N_MAX]={};
  scanf("%ld %ld %ld",&N,&K,&C);
  scanf("%s",S);
  for(long a=0;a<N;a++){
    if(magic==0){
      if(hantei(a,hako,K))continue;
    };
    kago=S[a];
    S[a]='x';
    m=0;
    while(m<K){
      if(p>N-1){
        printf("%ld\n",a+1);
        break;
      }
      else if(S[p]=='o'){
       if(magic)hako[m]=p;
        p+=C+1;
        m++;
      }
      else if(S[p]=='x'){
        p++;
      };
     // printf("bbb%ld\n",m);
    };
    if(magic)magic--;
    S[a]=kago;
    p=0;
  };
};