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<iostream>
#include<cstdio>
#include <stdio.h>
#include<algorithm>
#include<cstring>
#include <string>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<map>
#include<vector>
#include<bits/stdc++.h>
#include <set>
#define ll  unsigned long long
#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#define inf 0x3f3f3f3f
#define pi 3.1415926535898
using namespace std;
ll  a[100010],b[100010],c[100010];
ll  n;
ll  ans=0;
int main()
{

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