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 INF 5000000000000000000
#define ll long long
#define pll pair<ll, ll>
using namespace std;

int main() {
  ll N;
  cin >> N;
  vector<ll> A(N + 1);
  ll s = 0;
  for (ll i = 0; i <= N; ++i) {
    cin >> A.at(i);
    s += A.at(i);
  }
  ll cnt = 1;
  ll node = 1;
  ll ans = 0;
  for (ll i = 0; i <= N; ++i) {
    ans += node;
    if (node >= A.at(i)) {
      ll temp = min(s - cnt, node - A.at(i));
      cnt += temp;
      node = (node - A.at(i)) + temp;
    }
    else {
      cout << -1 << "\n";
      return 0;
    }
  }
  cout << ans << "\n";
}
