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"

func mini(x int, y int) int {
	if x < y {
		return x
	} else {
		return y
	}
}

func main(){
	var n,ans,t int
	fmt.Scan(&n)
	a := make([]int,n+1)
	capa := make([]int,n)
	temp := 1
	flag := true
	if n == 0{
		flag = false
	}
	k := 1
	for i:=0 ; i<n ; i++ {
		fmt.Scan(&a[i])
		capa[i] = temp - a[i]
//fmt.Println(i,capa[i])
		temp = capa[i]*2
		if capa[i] <= 0 || a[i] > k{
			flag = false
		}
		k *= 2
	}
	fmt.Scan(&ans)
p := make([]int,n)
//fmt.Println(capa)
	if !flag || ans > k{
		fmt.Println("-1")
	} else {
		child := ans
		for i:=n-1 ; i>=0 ; i-- {
			t = child - capa[i]*2
			if t > 0 {
				flag = false
				break
			}
			temp = mini(child,capa[i]) + a[i]
//fmt.Println(temp)
			ans += temp
			child = temp
			p[i] = temp
		}
		if flag {
			child = 1
			for i:=0 ; i<n ; i++ {
				if p[i] > child {
					flag = false
					break
				}
				child = (child-a[i])*2
			}
			if flag {
				fmt.Println(ans)
			} else {
				fmt.Println("-1")
			}
		} else {
			fmt.Println("-1")
		}
	}
}