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: Runtime Error)

n = int(input())
a = list(map(int,input().split()))
ma1 = [1]*n
ma2 = [0]*n
mi2 = [0]*n

for i in range(1,n):
    ma1[i]=ma1[i-1]*2-a[i]
    if ma1[i]<0:
        print(-1)
        exit()
if n == 1:
    if a[0]==1:
        print(-1)
        exit()
    if a[0]==0:
        if a[1]==1:
            print(2)
            exit()
        elif a[1]==2:
            print(3)
            exit()
        else:
            print(-1)
            exit()

ma2[n-1]=a[n]
ma2[n-2]=ma2[n-1]
mi2[n-1]=a[n]
mi2[n-2]=mi2[n-1]//2 + mi2[n-1]%2
if n!=2:
    for i in range(2,n):
        ma2[n-1-i]=min(ma2[n-i]+a[n-i+1],2**(n-i)-a[n-i])
        mi2[n-1-i]=(mi2[n-i]+a[n-i+1])//2 + (mi2[n-i]+a[n-i+1])%2
        if mi2[n-1-i]>ma2[n-1-i]:
            print(-1)
            exit()
ma2 = [1]+ma2
ans = 0
for i in range(n):
    ans += min(ma1[i],ma2[i])
ans +=sum(a)

print(ans)