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<iostream>
#define ll long long
#define P pari<int,int>
#define mk make_pair
const int N = 1e5+10;
using namespace std;
ll a[N],b[N];
int main()
{
    int n;
    cin >> n;
    for(int i=0;i<=n;i++)
    {
        cin >> a[i];
    }
    b[n] = a[n];
    for(int i=n-1;i>=0;i--)
    {
        if(i<=60)
        {
            b[i] = min(b[i+1] + a[i],1LL<<i);
        }
        else
            b[i] = b[i+1] + a[i];
    }
    ll ans = 0,temp = 1,flag = 1;
    for(int i=0;i<n;i++)
    {
        ans += temp;
        temp -= a[i];
        if(temp > b[i+1] || 2*temp < a[i+1] )
            flag = 0;
        temp = temp * 2;
        temp = min(temp,b[i+1]);
    }
    if(flag)
        cout << ans + b[n];
    else
        cout << "-1";
    return 0;
}