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)

import math

def read_int():
    return int(input())

def read_int_map():
    return map(int,input().split(' '))

def read_int_list():
    return list(map(int,input().split(' ')))

def read_s_list_loop(n):
    return [input() for _ in range(n)]


N = int(input())
An = read_int_list()

remain_trees = An[N]
amount = remain_trees
'''
tree = [ [] for _ in range(N)]
tree_count = 0
'''
threads_holds = [0]*(10**5+1)
threads_holds[0] = 1
tree = [0]*(N+1)
tree[N] = remain_trees
for i in range(1,10**5+1):
    threads_holds[i] = threads_holds[i-1]*2

for i in reversed(range(1,N)):
    shoud_end = An[i]
    
    #ありえない数なので減らす必要ある
    # 1/2以下 + should_endがthread_hold以上の時はprint -1
    if threads_holds[i] < int(math.ceil(remain_trees/2.0)) + shoud_end:
        print(-1)
        exit(0)
    else:
        if i != 0:
            next_should_end = An[i-1]
            max_val_for_next = (threads_holds[i-1] - next_should_end)*2
            if max_val_for_next < int(math.ceil(remain_trees/2.0)) + shoud_end:
                print(-1)
                exit(0)
            remain_trees = min(max_val_for_next,remain_trees + shoud_end) 
    amount += remain_trees
    tree[i] = remain_trees
if remain_trees <= (threads_holds[0] - An[0])*2:
    print(amount+1)
else:
    print(-1)