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 <cstdio>
#include <cstring>
#include <cmath>
#include <utility>
#include <iostream>
#include <functional>
#include <bitset>
#include <algorithm>
#include <vector>
#include <forward_list>
#include <set>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <tuple>
#include <numeric>
#define rep(i, s, g) for ((i) = (s); (i) < (g); ++(i))
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
const ll MOD = 1e9 + 7;
const ll INF = (1ll << 60);

struct tree
{
    ll root;
    ll edge;
};

int main(void)
{
    ll n;
    cin >> n;

    vector<ll> a(n + 3), pok(n + 3), v(n + 3);

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

    pok[0] = 1 - a[0];

    if (pok[0] < 0)
    {
        cout << -1 << endl;
        return 0;
    }

    for (ll i = 1; i <= n; i++)
    {
        pok[i] = pok[i - 1] * 2 - a[i];

        if (pok[i] < 0)
        {
            cout << -1 << endl;
            return 0;
        }
    }

    ll ans = 0;
    v[n] = a[n];

    for (ll i = n - 1; i >= 0; i--)
    {
        v[i] = min(v[i + 1], pok[i]) + a[i];
    }

    for (ll i = 0; i <= n; i++)
    {
        ans += v[i];
    }

    if (ans <= 0)
    {
        cout << -1 << endl;
        return 0;
    }

    cout << ans << endl;
}