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 NOMURA2020C
  integer(8)::N,i
  integer(8),allocatable::A(:),k(:),l(:)
  read(*,*)N
  allocate(A(0:N),k(0:N),l(0:N))
  read(*,*)A
  if(A(0)==0) then
     k(0)=1_8
  else
     k(0)=0_8
  end if
  do i=1,N
     k(i)=k(i-1)*2_8-A(i)
     if(k(i)<0) then
        write(*,*)'-1'
        stop
     end if
  end do
  l(N)=A(N)
  do i=N-1_8,0_8,-1_8
     l(i)=l(i+1)+A(i)
     if(l(i)>k(i)+A(i)) l(i)=k(i)+A(i)
  end do
  write(*,*) sum(l)
end program NOMURA2020C