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.

3
0 1 1 2

7

Below is the tree with the maximum possible number of vertices. It has seven vertices, so we should print $7$.

4
0 0 1 0 2

10

2
0 3 1

-1

1
1 1

-1

10
0 0 1 1 2 3 5 8 13 21 34

264