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 Fortran (GNU Fortran 9.2.1) (judge verdict: Accepted)

program main
    use,intrinsic :: iso_fortran_env
    implicit none
    integer(int64):: n,i,j,asum,pmax,ans=0
    integer(int64), allocatable:: a(:)

    read*, n
    allocate(a(0:n))
    read*, a(:)
    asum = sum(a)
    pmax = 1
    i=0
    do while(pmax < asum)
        ans = ans + min(pmax, asum)
        if (a(i) > pmax) then
            print'(i0)', -1
            stop
        end if
        asum=asum-a(i)
        pmax=(pmax-a(i))*2
        i=i+1
    end do

    do j=i,n
        ans = ans + asum
        asum=asum-a(j)
    end do
    print'(i0)', ans
end program main