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].
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF807E {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		StringTokenizer st = new StringTokenizer(br.readLine());
		int[] cnt = new int[42];
		int[] cnt1 = new int[42];
		int[] cnt2 = new int[42];
		for (int i = 0; i < n; i++) {
			long a = Long.parseLong(st.nextToken());
			int k = 0;
			while (1L << k < a)
				k++;
			if (1L << k == a)
				cnt[k]++;
			else
				cnt1[k]++;
		}
		for (int k = 1; k <= 40; k++) {
			int d = cnt[k] - cnt[k - 1];
			if (d > 0) {
				cnt1[k] += d;
				cnt[k] -= d;
			}
		}
		int m = cnt[0];
		int[] aa = new int[m];
		for (int h = 0, k = 0; k <= 40; k++)
			for (int i = 0; i < cnt[k] - cnt[k + 1]; i++)
				aa[h++] = k;
		int[] bb = new int[n];
		int lower = -1, upper = m + 1;
		while (upper - lower > 1) {
			int p = (lower + upper) / 2;
			for (int k = 0; k <= 40; k++)
				cnt2[k] = cnt1[k];
			for (int h = 0; h < p; h++)
				for (int k = aa[h]; k >= 0; k--)
					cnt2[k]++;
			int q = 0;
			for (int k = 0; k <= 40; k++)
				for (int i = 0; i < cnt2[k]; i++)
					bb[q++] = k;
			boolean yes = true;
			if (m - p < q)
				yes = false;
			else
				for (int i = m - 1, j = q - 1; j >= 0; i--, j--)
					if (aa[i] < bb[j] - 1) {
						yes = false;
						break;
					}
			if (yes)
				lower = p;
			else
				upper = p;
		}
		if (lower == -1)
			System.out.println(-1);
		else {
			StringBuilder sb = new StringBuilder();
			for (int p = lower; p >= 0; p--)
				sb.append(m - p + " ");
			System.out.println(sb);
		}
	}
}
