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: Accepted)

def Int():
    return int(input())
def Ints():
    return map(int,input().split())
def IntList():
    return list(Ints())
def IntMat(N):
    return [IntList() for i in range(N)]

import sys
sys.setrecursionlimit(4100000)
rl = sys.stdin.readline
mod = 10**9+7

N = Int()
A = IntList()

tmp = 0

for i in range(N+1):
    tmp += A[i]*2**(N-i)
    
if tmp > 2**N:
    print(-1)
    sys.exit()
    
if N==0:
  print(1)
  sys.exit()


B = A[::-1]
ans = 1
par = 1
C = []
MAXpar = B[0]

for i in range(N):
    leaf = B[i]
    if len(C)>=1:
        MINP = C[-1][0]
        MAXP = C[-1][1]
        C.append([(MINP+leaf+1)//2,MAXP+leaf])
    else:
        C.append([(leaf+1)//2,leaf])
        

D = C[::-1]
#print(D)
for i in range(1,N):
    minp,maxp = D[i]
    
    ans += A[i]+min(maxp,2*par-A[i])
    #print(A[i],min(maxp,2*par-A[i]))
    par = min(maxp,2*par-A[i])
    
ans += A[-1]
print(ans)