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 C (GCC 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
int main()
{
	int n;
	scanf("%d", &n);
	int i;
	long long int a[100005];
	for (i = 0; i <= n; i++)
		scanf("%lld", &a[i]);
	long long int max[100005], min[100005];
	max[n] = min[n] = a[n];
	for (i = n - 1; i >= 0; i--)
	{
		max[i] = max[i + 1] + a[i];
		min[i] = (min[i + 1] + 1) / 2 + a[i];
	}
	if (min[0] > 1)
	{
		printf("-1\n");
		return 0;
	}
	long long int ans = 0;
	long long int p = 1;
	for (i = 0; i <= n; i++)
	{
		ans += p;
		p = (p - a[i]) * 2;
		if (p > max[i + 1])
			p = max[i + 1];
	}
	printf("%lld\n", ans);
	return 0;
}