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;
using ll = long long;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll n;
    cin >> n;
    vector<ll> as(n+1);
    for(ll i=0; i<=n; i++)
    {
        cin >> as.at(i);
    }

    vector<ll> nemax(n+1);
    nemax.at(0) = 1;
    for(ll i=0; i<n; i++)
    {
        if(nemax.at(i) <= as.at(i))
        {
            cout << "-1\n";
            return 0;
        }
        nemax.at(i+1) = (nemax.at(i) - as.at(i)) * 2;
        if(nemax.at(i+1) >= static_cast<ll>(2e8+100)) nemax.at(i+1) = static_cast<ll>(2e8+100);
    }
    // for(auto ne : nemax) cout << ne << '\n';

    ll num = 0;
    ll ans = 0;
    for(ll i=n; i>=0; i--)
    {
        num += as.at(i);
        if(nemax.at(i) < num) num = nemax.at(i);
        ans += num;
    }
    cout << ans << '\n';

    return 0;
}