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;
#pragma comment(linker, "/STACK:102400000,102400000")
const long long INF = 1e18 + 1LL;
const double Pi = acos(-1.0);
const int N = 1e5 + 10, M = 1e3 + 20, mod = 1e9 + 7, inf = 2e9;
long long H[70], a[N];
int No, cnt[N], n, cnts;
vector<long long> G, ans;
vector<long long> all[N];
int sum[N], sum2[N];
pair<int, long long> P[N];
void go(long long x) {
  int i;
  for (i = 0; i <= 60; ++i) {
    if (x < H[i]) break;
  }
  i--;
  for (int j = 0; j <= i; ++j) {
    cnt[j]--;
    if (cnt[j] < 0) No = 1;
    return;
  }
}
int cango(long long x) {
  if (x == 0) return 0;
  int ok = 0;
  for (int i = 0; i <= 60; ++i) {
    if (H[i] <= x) {
      cnt[i]--;
      if (cnt[i] < 0) {
        ok = 1;
      }
    }
  }
  if (ok) {
    for (int i = 0; i <= 60; ++i)
      if (H[i] <= x) cnt[i]++;
    return 0;
  } else
    return 1;
}
int can(long long now) {
  for (int i = G.size() - 1; i >= 0; --i) {
    int ok = 0;
    for (int j = 1; j <= 60; ++j) {
      if (G[i] <= H[j] && sum[j - 1]) {
        sum[j - 1]--;
        P[++cnts] = make_pair(j - 1, G[i]);
        ok = 1;
        G.pop_back();
        break;
      }
    }
    if (!ok) return 0;
  }
  return 1;
}
int allcan(int x) {
  int j = x + 1, i = 0;
  int ok;
  while (j <= cnts && i < G.size()) {
    if (P[j].second != 0)
      j++;
    else if (H[P[j].first + 1] < G[i])
      j++;
    else
      i++, j++;
  }
  if (i == G.size()) {
    return 1;
  } else
    return 0;
}
int check(int x) {
  x = cnts - x;
  if (x > cnts) return 0;
  if (x == 0) return 1;
  G.clear();
  for (int i = 1; i <= x; i++) {
    for (int j = 0; j <= P[i].first; ++j) {
      G.push_back(H[j]);
    }
    if (P[i].second) {
      G.push_back(P[i].second);
    }
  }
  if (allcan(x)) {
    return 1;
  } else
    return 0;
}
int main() {
  H[0] = 1;
  for (int i = 1; i <= 60; ++i) H[i] = H[i - 1] * 2LL;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", &a[i]);
    int ok = 0;
    for (int j = 0; j <= 60; ++j) {
      if (a[i] == H[j]) {
        ok = 1;
        cnt[j]++;
        break;
      }
    }
    if (!ok) G.push_back(a[i]);
  }
  int now = 0;
  for (int i = 60; i >= 0; --i) {
    while (cnt[i]) {
      if (cango(H[i])) {
        now++;
        sum[i]++;
      } else
        break;
    }
  }
  for (int i = 0; i <= 60; ++i)
    for (int j = 1; j <= cnt[i]; ++j) G.push_back(H[i]);
  int l = 1, r, ans = -1, tmpr;
  if (can(now))
    r = now;
  else
    r = -1;
  tmpr = r;
  for (int i = 0; i <= 60; ++i) {
    for (int j = 1; j <= sum[i]; ++j) {
      P[++cnts] = make_pair(i, 0);
    }
  }
  sort(P + 1, P + cnts + 1);
  while (l <= r) {
    int md = (l + r) >> 1;
    if (check(md)) {
      ans = md;
      r = md - 1;
    } else
      l = md + 1;
  }
  if (tmpr == -1)
    puts("-1");
  else {
    for (int i = ans; i <= tmpr; ++i) cout << i << " ";
    cout << endl;
  }
  return 0;
}
