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>

using namespace std;

typedef long long ll;

const ll INF = 1e17;

ll t[100005];
ll a[100005];
ll ans;
int main()
{
    int n;
    cin >> n;

    for (int i = 0; i <= n; i++)
    {
        cin >> a[i];
    }
    t[0] = 1;

    if (n == 0 && a[0] == 1)
    {
        cout << 1 << endl;
        return 0;
    }

    if (a[0] != 0)
    {
        cout << -1 << endl;
        return 0;
    }
    for (int i = 1; i <= n; i++)
    {
        t[i] = t[i - 1] * 2;
        if (t[i] >= INF)
            t[i] = INF;
        t[i] -= a[i];
        if (t[i] < 0)
        {
            cout << -1 << endl;
            return 0;
        }
    }

    ll m = a[n];
    ans = m;
    for (int i = n; i > 0; i--)
    {

        m = min(m, t[i - 1]);
        m += a[i - 1];
        ans += m;
    }

    cout << ans << endl;
}