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)

n=int(input())
a=list(map(int,input().split()))
if n==0 and a==[1]:
	print(1)
	exit()
elif n==0 and a!=[1]:
	print(-1)
	exit()
elif a[0]==1:
	print(-1)
	exit()
else:
	ruisekiwa = [0 for _ in range(n+1)]
	ruisekiwa[n] = a[n]
	for i in range(n)[::-1]:
		ruisekiwa[i] = a[i] + ruisekiwa[i+1]
	cnt=1
	ans=1
	for i in range(1,n+1):
		target = a[i]
		if cnt*2 - target <= 0 and i != n:
			print(-1)
			exit()
		else:
			if i == n:
				if cnt * 2 - target < 0:
					print(-1)
					exit()
			if cnt*2-a[i] < ruisekiwa[i]:
				ans += cnt*2
				cnt = cnt*2-a[i]
			else:
				ans += ruisekiwa[i]
				cnt = ruisekiwa[i]
	print(ans)