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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define pb push_back
#define mp make_pair
#define F  first
#define S  second
#define ff(i,n) for(i=0;i<n;i++)
ll bg=9982443989853;
ll power(ll x,ll y)
{
	if(y==0)
		return 1;
	ll h=power(x,y/2)%bg;	
	h=(h*h)%bg;
	if(y%2==1)
	  return (x*h)%bg;
	else
	  return h%bg;  
}
ll modinv(ll m)
{
	return power(m,bg-2)%bg;
}
ll arr[600000];
string f[20];
ll check(string a,ll n)
{
	ll heck = 0,i,j,m=a.length();
	ff(i,n)
		{
			ll c=0;
			ff(j,m)
			  if(a[j]!=f[i][j])
			     c++;
			if(c>1)
			   heck =1;     
		}
	return heck ;	
}
int main()
{ 
	ll t;
	t=1;
	while(t--)
	{
		ll n,i,vertex = 0;
		cin>>n;
		ll a[n+1],b[n+1];
		ff(i,n+1)
		   cin>>a[i];
		b[0] = 1-a[0];
		if(b[0]<0)
		{
			cout<<"-1";
			return 0;	
		}   
		for(i=1;i<=n;i++)
		{
			b[i] = 2*b[i-1]-a[i];
			if(b[i]<0)
			{
				cout<<"-1";
				return 0;
			}
		}
		for(i=n-1;i>=0;i--)
			b[i]=min(b[i],a[i+1]+b[i+1]);
		for(i=0;i<n;i++)
			vertex+=(a[i]+b[i]);
		vertex+=a[n];
		cout<<vertex;			
	}   
	return 0;
}