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: Wrong Answer)

program folia
  implicit none
  integer :: n, i
  integer(8) :: a(0:100001) = 0, p(0:60) = 0, r, m = 0
  read(*,*) n
  read(*,*) a(0:n)
  p = (/(ibset(0_8, i), i = 0, 60)/)
  do i = min(60, n), 0, -1
    r = (a(i + 1) + 1) / 2 + a(i)
    if (r > p(i)) then
      write(*,'(i0)') -1
      stop
    end if
  end do
  r = 0
  do i = n, 61, -1
    m = m + a(i)
    r = r + m
  end do
  do i = min(n, 60), 1, -1
    m = min(m + a(i), (p(i - 1) - a(i - 1)) * 2)
    r = r + m
  end do
  write(*,'(i0)') r + 1
end program folia