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>
int main(void){
    // Your code here!
    int n,m,x,j,count;
    char s[200100]={};
    char w1[200100]={};
    char w2[200100]={};
    scanf("%d %d %d %s",&n,&m,&x,s);
    count = 0;
    for(j=0;j<n;j++){
        if(s[j]=='o'){
            w1[count]=j;
            //printf("w1 %d\n",j);
            count=count+1;
            j = j + x;
        }
        if(count == m){ break;}
    }
    count = 0;
    for(j=0;j<n;j++){
        if(s[n-j-1]=='o'){
            w2[count]=(n-j-1);
            //printf("w2 %d\n",n-j-1);
            count=count+1;
            j = j + x;
        }
        if(count == m){ break;}
    }
    for(j=0;j<m;j++){
        if( w1[j]==w2[m-j-1]){
                //printf("j%d %d %d %d\n",j,w1[j]+1,m-j-1,w2[m-j-1]);
                printf("%d\n",w1[j]+1);
        }
    }
    return 0;
}