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"
	"os"
	"strconv"
)

const (
	initialBufSize = 100000
	maxBufSize     = 10000000
)

var (
	sc = bufio.NewScanner(os.Stdin)
	wt = bufio.NewWriter(os.Stdout)
)

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

func nextInt() int {
	i, _ := strconv.Atoi(next())
	return i
}

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

func puts(a ...interface{}) {
	fmt.Fprintln(wt, a...)
}

func min(nums ...int) int {
	ret := nums[0]
	for _, v := range nums {
		if v < ret {
			ret = v
		}
	}
	return ret
}

func main() {
	sc.Split(bufio.ScanWords)
	sc.Buffer(make([]byte, initialBufSize), maxBufSize)
	defer wt.Flush()

	n := nextInt()
	a := nextInts(n + 1)

	l, r := make([]int, n+1), make([]int, n+1)
	l[n], r[n] = a[n], a[n]
	for i := n - 1; i >= 0; i-- {
		l[i] = a[i] + (l[i+1]+1)/2
		r[i] = a[i] + r[i+1]
	}
	if l[0] > 1 {
		puts(-1)
		return
	}

	ans, cur := 0, 1
	for i := 0; i <= n; i++ {
		ans += cur
		if i < n {
			cur = min((cur-a[i])*2, r[i+1])
		}
	}
	puts(ans)
}