It can be shown that any positive integer x can be uniquely represented as x = 1 + 2 + 4 + ... + 2k - 1 + r, where k and r are integers, k ≥ 0, 0 < r ≤ 2k. Let's call that representation prairie partition of x.

For example, the prairie partitions of 12, 17, 7 and 1 are: 

12 = 1 + 2 + 4 + 5,

17 = 1 + 2 + 4 + 8 + 2,

7 = 1 + 2 + 4,

1 = 1. 

Alice took a sequence of positive integers (possibly with repeating elements), replaced every element with the sequence of summands in its prairie partition, arranged the resulting numbers in non-decreasing order and gave them to Borys. Now Borys wonders how many elements Alice's original sequence could contain. Find all possible options!

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the number of numbers given from Alice to Borys.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1012; a1 ≤ a2 ≤ ... ≤ an) — the numbers given from Alice to Borys.

Output

Output, in increasing order, all possible values of m such that there exists a sequence of positive integers of length m such that if you replace every element with the summands in its prairie partition and arrange the resulting numbers in non-decreasing order, you will get the sequence given in the input.

If there are no such values of m, output a single integer -1.

Examples

Input

8
1 1 2 2 3 4 5 8


Output

2 


Input

6
1 1 1 2 2 2


Output

2 3 


Input

5
1 2 4 4 4


Output

-1

Note

In the first example, Alice could get the input sequence from [6, 20] as the original sequence.

In the second example, Alice's original sequence could be either [4, 5] or [3, 3, 3].
import java.util.Scanner;


public class E7 {
	
	@SuppressWarnings("resource")
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		int n = input.nextInt();
		int[] powers = new int[43];
		int[] lesser = new int[43];
		long power = 1;
		int index = 0;
		for (int i = 0; i < n; i++){
			long num = input.nextLong();
			while(num > power){
				power *= 2;
				index++;
			}
			if (num == power){
				powers[index]++;
			}else{
				lesser[index]++;
			}
		}
		boolean printed = false;
		for (int seq = 1; seq <= powers[0]; seq++){
			int running = seq;
			int ends = 0;
			for (index = 0; index < 42; index++){
				ends += lesser[index];
				if (powers[index] > running){
					ends += powers[index] - running;
					//System.out.println(index + "A" + running + ends);
				}else{
					int stopping = running - powers[index];
					ends -= Math.min(stopping, ends);
					running = powers[index];
					//System.out.println(index + "B" + running + ends);
				}
			}
			if (ends == 0){
				if(printed){
					System.out.print(" ");
				}
				System.out.print(seq);
				printed = true;
			}
		}
		if(!printed){
			System.out.println(-1);
		}else{
			System.out.println();
		}
	}

}
