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;
typedef long long ll;
//cout << fixed << setprecision(17) << res << endl;                                                    
int main() {
  ll n;
  cin >> n;
  vector<ll> A(n+1);
  for (auto & a : A) cin >> a;

  ll total = 0;
  for (auto a : A) total += a;


  ll res = 0;
  ll cur = 1;
  for (ll i = 0; i <= n; ++i) {
    if (cur < A[i]) {
      cout << -1 << endl;
      return 0;
    }
    res += min(cur, total);
    cur = min(cur, total) - A[i];
    total -= A[i];
    cur *= 2;
  }
  cout << res << endl;
  return 0;
}