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 Go (1.14.1) (judge verdict: Wrong Answer)

package main

import (
	"fmt"
	"math"
)

func main() {
	var N int
	var a int
	fmt.Scanf("%d", &N)
	A := make([]int, N+1)
	for i := 0; i <= N; i++ {
		fmt.Scanf("%d", &a)
		A[i] = a
	}

	var maxLeaves float64
	maxLeaves = math.Pow(2.0, float64(N))

	out := math.Pow(2, float64(N+1)) - 1
	for i := 1; i < N; i++ {
		// 頂点減らす
		out -= math.Pow(2.0, float64(N-i+1))*float64(A[i]) - 2.0
		// 可否判断
		maxLeaves -= math.Pow(2.0, float64(N-i)) * float64(A[i])

		// 木が作れないなら終わり
		if maxLeaves < float64(A[N]) {
			fmt.Println("-1")
			return
		}
	}
	fmt.Println(out)
}