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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.HashMap;
import java.util.Scanner;

public class Main {
	public static void main(String[] args){
		try(Scanner sc = new Scanner(System.in)) {

			int n = sc.nextInt();
			int k = sc.nextInt();
			int c = sc.nextInt();
			String s = sc.next();

			//もともとoである対象日をxとした場合に答えが求められるかどうかで決定する
			//充足探索関数を作る
			for(int i = 0 ; i < n ; i++ ) {
//				System.out.println(i+1);
				if ( !canWork(k , c , s, i, 0)) {
					System.out.println(i+1);
				}
			}
		}	
	}
	
	private static HashMap<String, Boolean> cache = new HashMap<>();

	private static boolean canWork(int k, int c, String s, int target, int current) {

		if ( k == 0 ) {
			return true;
		}
		//find first o and (not target) from current
		while(current < s.length()) {
			if ( current > target && cache.containsKey(current + "_" + k) ) {
//				System.out.println("## current = "+ current + ", " + cache.get(current + "_" + k));
				return cache.get(current + "_" + k);
			}
			if ( s.charAt(current) == 'o' && target != current ) {
//				System.out.println(String.format("## %d, %d, %d", k, target, current));
				//find first work;
				
				//find next
				int baseCurrent = current;
				current += c+1;
				boolean result = canWork(k-1, c, s, target, current);
				
				if ( baseCurrent > target ) {
					cache.put(baseCurrent + "_" + k, result);
				}
				
				return result;
			}
			current++;
		}
		return false;
	}


}