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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<iostream>
#include<cstring>
using namespace std;
typedef long long LL;
const int N = 1e5;
LL a[N+1];
LL bck[N+1];
LL p[N+1];
LL solve(LL a[], int n)
{
	if(a[0] > 0) //and depth >= 1
		return -1;
	bck[n] = a[n];
	for(int i = n-1; i >= 0; i--)
		bck[i] = bck[i+1]+a[i];

	p[0] = 1;
	for(int i = 1; i < n; i++)
	{
		p[i] = min(2*p[i-1]-a[i],bck[i+1]);
		if(!(p[i-1] <= p[i]+a[i]))
			return -1;
	}
	p[n] = 0;

	LL ans = 0;
	for(int i = 0; i <= n; i++)
		ans += a[i]+p[i];
	return ans;
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	int n; cin >> n;
	for(int i = 0; i <= n; i++)
		cin >> a[i];
	cout<<solve(a,n)<<endl;

	cout<<flush;
	return 0;
}