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].
#include <bits/stdc++.h>
long long arr[100001];
std::map<long long, int> amt;
std::vector<std::pair<long long, int>> removed;
void restoreMap() {
  for (auto it = removed.begin(); it != removed.end(); it++) {
    amt[it->first] += it->second;
  }
  removed.clear();
}
int test(int max) {
  std::queue<std::pair<long long, int>> q;
  int curr = max;
  long long val = 1;
  int used = -1;
  while (curr) {
    auto it = amt.find(val);
    int take = 0;
    if (it != amt.end()) {
      take = std::min(curr, it->second);
      it->second -= take;
      removed.push_back(std::make_pair(it->first, take));
    }
    if (curr < 1e9 && curr - take > 0) {
      q.push(std::make_pair(val, curr - take));
    }
    curr = take;
    val *= 2;
    if (used == -1) used = take;
  }
  auto it = amt.begin();
  if (it == amt.end()) {
    restoreMap();
    return used;
  }
  int canTake = it->second;
  while (!q.empty()) {
    std::pair<long long, int> p = q.front();
    q.pop();
    while (true) {
      if (it->first > p.first) {
        break;
      }
      int take = std::min(canTake, p.second);
      canTake -= take;
      p.second -= take;
      if (canTake == 0) {
        it++;
        if (it == amt.end()) {
          restoreMap();
          return used;
        }
        canTake = it->second;
      }
      if (p.second == 0) {
        break;
      }
    }
  }
  while (canTake == 0) {
    it++;
    if (it == amt.end()) {
      restoreMap();
      return used;
    }
    canTake = it->second;
  }
  restoreMap();
  return -1;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &arr[i]);
    amt[arr[i]]++;
  }
  int max = test(1e9);
  if (max == -1) {
    printf("-1\n");
    return 0;
  }
  int min = -1;
  int low = 0, high = max;
  while (low <= high) {
    int mid = (low + high) / 2;
    int result = test(mid);
    if (result != -1 && result != mid) throw "";
    if (result == -1) {
      low = mid + 1;
    } else {
      min = mid;
      high = mid - 1;
    }
  }
  if (min == -1) throw "";
  for (int i = min; i <= max; i++) {
    printf("%d ", i);
  }
  printf("\n");
  return 0;
}
