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: Accepted)

package main

import (
	"bufio"
	"fmt"
	"log"
	"math"
	"os"
	"strconv"
)

var sc = bufio.NewScanner(os.Stdin)
var buff []byte

func init() {
	sc.Split(bufio.ScanWords)
	sc.Buffer(buff, bufio.MaxScanTokenSize*1024)
	log.SetFlags(log.Lshortfile)
}

var N int
var A []int
var ans int

func main() {
	N = nextInt()
	N++
	A = nextInts(N)
	if N == 0 {
		fmt.Println("0")
		return
	}
	if N == 1 {
		if A[0] == 1 {
			fmt.Println("1")
		} else {
			fmt.Println("-1")
		}
		return
	}
	if A[0] != 0 {
		fmt.Println("-1")
		return
	}
	L := make([]int, N)
	R := make([]int, N)
	for i := N - 2; i >= 0; i-- {
		L[i] = (A[i+1] + L[i+1] + 1) / 2
		R[i] = A[i+1] + R[i+1]
	}
	if !(L[0] <= 1 && 1 <= R[0]) {
		fmt.Println("-1")
		return
	}
	x := make([]int, N)
	x[0] = 1
	for i := 1; i < N; i++ {
		left := max(x[i-1]-A[i], L[i])
		right := min(2*x[i-1]-A[i], R[i])
		if left <= right {
			x[i] = right
		} else {
			fmt.Println("-1")
			return
		}
	}
	ans := 0
	for i := 0; i < N; i++ {
		ans += A[i] + x[i]
	}
	fmt.Println(ans)
}

func nextString() string {
	sc.Scan()
	return sc.Text()
}

func nextFloat64() float64 {
	sc.Scan()
	f, err := strconv.ParseFloat(sc.Text(), 64)
	if err != nil {
		panic(err)
	}
	return f
}

func nextInt() int {
	sc.Scan()
	i, err := strconv.Atoi(sc.Text())
	if err != nil {
		panic(err)
	}
	return i
}

func nextInts(n int) (r []int) {
	r = make([]int, n)
	for i := 0; i < n; i++ {
		r[i] = nextInt()
	}
	return r
}

var dy = []int{0, 1, 0, -1}
var dx = []int{1, 0, -1, 0}
var MAX = math.MaxInt32

func pow(a, b int) int {
	return int(math.Pow(float64(a), float64(b)))
}

func lcm(a, b int) int {
	return a * b / gcd(a, b)
}
func gcd(a, b int) int {
	if a < b {
		a, b = b, a
	}
	for b != 0 {
		a, b = b, a%b
	}
	return a
}
func max(a ...int) int {
	r := a[0]
	for i := 0; i < len(a); i++ {
		if r < a[i] {
			r = a[i]
		}
	}
	return r
}
func min(a ...int) int {
	r := a[0]
	for i := 0; i < len(a); i++ {
		if r > a[i] {
			r = a[i]
		}
	}
	return r
}
func sum(a []int) (r int) {
	for i := range a {
		r += a[i]
	}
	return r
}
func abs(a int) int {
	if a < 0 {
		return -a
	}
	return a
}

type Pair struct {
	a, b int
}

type Pairs []Pair

func (p Pairs) Len() int {
	return len(p)
}
func (p Pairs) Swap(i, j int) {
	p[i], p[j] = p[j], p[i]
}

func (p Pairs) Less(i, j int) bool {
	if p[i].a == p[j].a {
		return p[i].b < p[j].b
	}
	return p[i].a < p[j].a
}