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 Python (3.8.2) (judge verdict: Wrong Answer)

from sys import stdin
from sys import setrecursionlimit
from itertools import accumulate
setrecursionlimit(10 ** 7)

n = int(stdin.readline().rstrip())
a = list(map(int,stdin.readline().rstrip().split()))
b = list(accumulate(a[::-1]))
b = b[::-1]
#print(b)

if a[0] != 0:
    print(-1)
    exit()

ne = [0]*(n+1)

ne[0] = 1
point = 1

for i in range(1,n):
    point_ma = min(ne[i-1]*2,b[i+1]+a[i])
    if a[i] > point_ma:
        print(-1)
        exit()
    ne[i] = point_ma-a[i]
    point += point_ma

if ne[n-1]*2 < a[n]:
    print(-1)
    exit()

point += a[n]
ne[n] = 0

print(point)