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>
using namespace std;
const int maxm = 110;
int n;
long long x;
int ones;
int cnt[maxm];
int use[maxm];
int chain[maxm];
void find(const long long w) {
  int k = 0;
  long long e = 1;
  while (1) {
    if (w < e) {
      ++cnt[k - 1];
      break;
    } else if (w == e) {
      ++cnt[k];
      break;
    }
    k += 2;
    e <<= 1;
  }
}
bool solve(const int m) {
  memset(chain, 0, sizeof(chain));
  memcpy(use, cnt, sizeof(cnt));
  for (int k = 0; k < m; ++k) {
    int p = 0;
    while (use[p]) {
      --use[p];
      p += 2;
    }
    if (p) ++chain[p];
  }
  ones = use[0];
  for (int k = (maxm)-1; k >= 0; --k) {
    while (use[k]--) {
      bool finish = 0;
      for (int i = (maxm)-1; i >= 0; --i) {
        if (i < k) break;
        if (chain[i]) {
          finish = 1;
          --chain[i];
          break;
        }
      }
      if (!finish) return 0;
    }
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (n); ++i) {
    scanf("%I64d", &x);
    find(x);
  }
  vector<int> ans;
  int l = (cnt[0] + 1) >> 1, r = cnt[0];
  while (l <= r) {
    int m = (l + r) >> 1;
    if (solve(m)) {
      r = m - 1;
    } else {
      l = m + 1;
    }
  }
  if (solve(l)) {
    for (int i = (0); i <= (ones); ++i) ans.push_back(l + i);
  }
  if (ans.empty()) {
    printf("-1\n");
  } else {
    printf("%d", ans[0]);
    for (int i = 1; i < ((int)(ans).size()); ++i) {
      printf(" %d", ans[i]);
    }
    printf("\n");
  }
  return 0;
}
