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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

from sys import stdin, stdout
n = int(stdin.readline())
nmleaves = list(map(int, stdin.readline().strip().split()))

# pref[i]: sum of all elements in nmleaves after i, i.e. sum(nmleaves[i+1:])
full = sum(nmleaves)
tot = 0
pref = [0]*(n+1)
for i, l in enumerate(nmleaves):
    tot += l
    pref[i] = full - tot

if n==0:
    print(0)

else:
    if nmleaves[0]!=0:
        print(-1)
        exit()
    
    ## it exists
    ans = 0
    for level, leaves in enumerate(nmleaves):
        if leaves > 2**(level):
            print(-1)
            exit()

        if level == 0:
            ans += 1
            validnom = 1
        else:
            maxup = 2*validnom
            maxnom = maxup - leaves

            if level+1<len(nmleaves):
                validnom = min(maxnom, pref[level], 2**level)
                ans += (leaves + validnom)
            else:
                ans += leaves
    print(ans)