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>
#define N_MAX 200000


int step1(int num,char s[],int k,int Hako[],int c,int n){
  int m=0,p=0,b=0;
  char hosyu=s[num];
  s[num]='x';
  while(m<=k+1){
    if(m==k+1)b=2;
    else if(m==k)if(p>n-1)b=1;
    if(p>n-1){
        break;
      };
    while(s[p]=='x')p++;
    if(p<=n-1){
      Hako[m]=p;
      m++;
      p+=c+1;
    };
    //printf("%dを検査\n",num+1);
  };
  s[num]=hosyu;
  return b;
};


void step2(int num,char s[],int k,int c,int n){
  int m=0,p=0;
  char hosyu=s[num];
  s[num]='x';
  while(m<k){
    if(p>n-1){
        printf("%d\n",num+1);
        break;
      }
    while(s[p]=='x')p++;
    if(p<=n-1){
      m++;
      p+=c+1;
    };    
  };
  s[num]=hosyu;
};

int main(){
  int l=0;
  int N,K,C,p=0;
  int hako[N_MAX]={};
  char S[N_MAX]={};
  scanf("%ld %ld %ld",&N,&K,&C);
  scanf("%s",S);
  do{l=step1(p,S,K,hako,C,N);
     p++;
    }while(l==0);
  if(l!=2)for(l=0;l<K;l++)step2(hako[l],S,K,C,N);
};