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 JavaScript (Node.js 12.16.1) (judge verdict: Wrong Answer)

console.log((args=>{
  const [[N],A] = args.trim().split('\n').map( r=>r.split(' ').map(v=>v|0));
  const a = [1-A[0]];
  for ( let i = 1; i < N; i++ ) a.push(a[i-1] * 2 - A[i]);
  a.push(0);
  let n = A[N];
  for ( let i = N-1; i >= 0; i-- ) {
    const c = a[i+1] + A[i+1];
    if ( a[i] > c ) a[i] = c;
    else if ( a[i] * 2 < c ) return '-1';
    n += a[i] + A[i];
  }
  return `${n}`;
})(require('fs').readFileSync('/dev/stdin', 'utf8')));