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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define rep(i,l,n) for(int (i)=(l);(i)<(n);(i)++)
#define min(a,b) (a)<(b)?(a):(b)
#define max(a,b) (a)>(b)?(a):(b)
#define abs(a) (a)>0?(a):-(a)
#define ll long long

int main(){
	int N,K,C; scanf("%d %d %d",&N,&K,&C);
	char S[N+2]; scanf("%s",S);
	int fast[K],last[K];
	int n=0;
	for(int i=0;i<K;){
		if( n >= N ) break;
		//printf("judge %d ",n);
		if( S[n]!='x' ){
			fast[i]=n;
			n += C+1;
			i++;
			//printf("Yes\n");
		}else{
			n++;
			//printf("No\n");
		}
	}
	n = N-1;
	for(int i=0;i<K;){
		if( n<0 ) break;
		if( S[n]!='x' ){
			last[K-1-i]=n;
			n -= C+1;
			i++;
		}else{
			n--;
		}
	}
	//rep(i,0,K) printf("%d %d\n",fast[i],last[i]);
	rep(i,0,K) if( fast[i]==last[i] ) printf("%d\n",fast[i]+1);
	return 0;
}