Problem:
Score : 600 points

Problem Statement

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.

Notes

A binary tree is a rooted tree such that each vertex has at most two direct children.

A leaf in a binary tree is a vertex with zero children.

The depth of a vertex v in a binary tree is the distance from the tree's root to v. (The root has the depth of 0.)

The depth of a binary tree is the maximum depth of a vertex in the tree.

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.

Sample Input 1

3
0 1 1 2

Sample Output 1

7

Below is the tree with the maximum possible number of vertices. It has seven vertices, so we should print 7.

Sample Input 2

4
0 0 1 0 2

Sample Output 2

10

Sample Input 3

2
0 3 1

Sample Output 3

-1

Sample Input 4

1
1 1

Sample Output 4

-1

Sample Input 5

10
0 0 1 1 2 3 5 8 13 21 34

Sample Output 5

264

Solution in Perl (5.26.1) (judge verdict: Wrong Answer)

my @arr = <STDIN>;
@arr = map { chomp; [ split(/\s+/, $_) ] } @arr;
print main(@arr);

sub main {
    my ($n) = shift->@*;
    my (@arr) = shift->@*;
    my @arr2 = @arr;
    for my $i (0..$n) {
        $arr[$i] = $arr[$i]/(2**$i);
    }
    my $sum = 0;
    $sum += $_ for map { $_ } @arr;
    return -1 if $sum != 1;
    $sum = 0;
    for my $i (reverse 1..$n) {
      $sum += $arr2[$i];
      $arr2[$i-1] += $arr2[$i]/2;
    }
    return $sum + 1;
}