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 uint = unsigned int;

int main() {
  int n;
  cin >> n;
  n++;
  ll nnum = 1;
  ll fsum = 0;
  ll sum = 0;
  bool exist = true;
  vector<int> a(n);
  rep(i, n) {
    cin >> a[i];
    fsum += a[i];
  }
  rep(i, n) {
    if (!exist) continue;
    sum += nnum;
    nnum -= a[i];
    if (nnum < 0) exist = false;
    fsum -= a[i];
    if (nnum * 2 <= fsum)
      nnum *= 2;
    else
      nnum = fsum;
  }
  if (!exist)
    cout << -1 << endl;
  else
    cout << sum << endl;
  return 0;
}
