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.

Sample Input 1Copy

3
0 1 1 2

Sample Output 1Copy

7

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

Sample Input 2Copy

4
0 0 1 0 2

Sample Output 2Copy

10

Sample Input 3Copy

2
0 3 1

Sample Output 3Copy

-1

Sample Input 4Copy

1
1 1

Sample Output 4Copy

-1

Sample Input 5Copy

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

Sample Output 5Copy

264