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>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define pi 3.141592653589793238
#define int long long
using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

int mx[100005];
int tot[100005];

signed main()
{

ios_base::sync_with_stdio(false);
cin.tie(NULL);  
cout.tie(0);   

#ifndef ONLINE_JUDGE
    if(fopen("INPUT.txt","r"))
    {
    freopen ("INPUT.txt" , "r" , stdin);
    freopen ("OUTPUT.txt" , "w" , stdout);
    }
#endif  

    int n;
    cin>>n;
    int a[n+1];
    for(int i=0;i<=n;i++)
    {
        cin>>a[i];
    }
    mx[0]=1-a[0];
    if(mx[0]<0)
    {
        cout<<-1;
        return 0;
    }
    for(int i=1;i<=n;i++)
    {
        int z=mx[i-1]*2-a[i];
        if(z<0)
        {
            cout<<-1;
            return 0;
        }
        mx[i]=z;
    }
    tot[n]=a[n];
    for(int i=n-1;i>=0;i--)
    {
        int z=tot[i+1];
        tot[i]=a[i]+min(mx[i],tot[i+1]);
    }
    for(int i=1;i<=n;i++)
    {
        int z=tot[i-1]-a[i-1];
        z*=2;
        if(tot[i]>z)
        {
            cout<<-1;
            return 0;
        }
    }
    int ans=0;
    for(int i=0;i<=n;i++)
    {
        ans+=tot[i];
    }
    cout<<ans;
}