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 <stdio.h>
typedef long long ll;

int main(void) {
  ll i, j, k, n, now = 1, ans;
  scanf("%lld", &n);
  ll a[n + 1], b[n + 1];
  for(i = 0; i <= n; ++i) {
    scanf("%lld", &a[i]);
    now -= a[i];
    if(now < 0) {
      printf("-1");
      return 0;
    }
    b[i] = now;
    now <<= 1ll;
    if(now > 1e16) now = 1e16;
  }
  now = ans = a[n];
  for(i = n; i; --i) {
    now = now < b[i - 1] ? now : b[i - 1], now += a[i - 1];
    ans += now;
  }
  printf("%lld", ans);
  return 0;
}