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 <iostream>
#include <vector>
using namespace std;

int main() {
  int n;
  scanf("%d", &n);
  vector<int> a(n + 1);
  for (int i = 0; i <= n; i++) scanf("%d", &a[i]);

  vector<long long> s(n + 1);
  for (int i = n - 1; i >= 0; i--) s[i] = s[i + 1] + a[i + 1];

  long long ans = 0, now = 1;
  bool f = true;
  for (int i = 0; i <= n; i++) {
    if (now < a[i]) {
      f = false;
      break;
    }

    now -= a[i];
    if (now > s[i]) now = s[i];
    ans += a[i] + now;
    now <<= 1;
  }

  if (f)
    printf("%lld\n", ans);
  else
    printf("-1\n");

  return 0;
}
