Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.

Constraints

* 0 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{8} (0 \leq i \leq N)
* A_N \geq 1
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_0 A_1 A_2 \cdots A_N


Output

Print the answer as an integer.

Examples

Input

3
0 1 1 2


Output

7


Input

4
0 0 1 0 2


Output

10


Input

2
0 3 1


Output

-1


Input

1
1 1


Output

-1


Input

10
0 0 1 1 2 3 5 8 13 21 34


Output

264
#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

int main() {
  int n;
  cin >> n;
  vector<int> a(n+1);
  rep(i, n+1) cin >> a[i];

  vector<ll> cum(n+1);
  cum[n] = a[n];
  for (int i = n-1; i >= 0; i--) {
    cum[i] = a[i] + cum[i+1];
  }

  ll nd = 1;
  ll ans = 0;
  rep(i, n+1) {
    if (nd < a[i]) {
      ans = -1;
      break;
    }
    ans += nd;
    nd -= a[i];
    if (i < n) nd = min(2*nd, cum[i+1]);
    // printf("ans = %lld, nd = %lld\n", ans, nd);
  }
  cout << ans << endl;
  return 0;
}