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;

long long n;
vector<long long> a, memo;

long long solve();

int main() {
  cin >> n;
  ++n;
  a.resize(n);
  for (auto &p : a) cin >> p;
  cout << solve() << endl;
  return 0;
}

long long solve() {
  long long res = 0, now = 1;
  memo.assign(n, 0);
  memo[n - 1] = a[n - 1];
  for (int i = n - 2; i >= 0; --i) memo[i] = memo[i + 1] + a[i];
  now = 1;
  for (int i = 0; i < n; ++i) {
    now = min(memo[i], now);
    res += now;
    now -= a[i];
    if (now < 0) return -1;
    now *= 2;
  }
  if (n == 1 && a[0] != 1) return -1;
  return res;
}
