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>
using namespace std;

int main()
{
  int64_t N, ans = 1;
  cin >> N;
  vector<int64_t> A(N + 1), MA(N + 1), MI(N + 1);
  for (int i = 0; i <= N; i++)
  {
    cin >> A.at(i);
  }
  MA[N] = MI[N] = A[N];
  for (int i = N - 1; i >= 0; i--)
  {
    MI[i] = (MI[i + 1] + 1) / 2 + A[i];
    MA[i] = MA[i + 1] + A[i];
  }
  if (!(MI[0] <= 1 && 1 <= MA[0]))
  {
    cout << -1 << endl;
    return 0;
  }
  int64_t prev = 1;
  for (int i = 1; i <= N; i++)
  {
    ans += min((prev - A[i - 1]) * 2, MA[i]);
    prev = min((prev - A[i - 1]) * 2, MA[i]);
  }
  cout << ans << endl;
}