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>
#include <string>
#include <vector>
#include <set>
#include <bits/stdc++.h>
#include <iomanip>
#include <numeric>
#include <utility>
#include <map>
#include <algorithm>
using namespace std;

int main(void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    long long N;
    cin >> N;
    vector<int> vec(N + 1);
    for (int i = 0; i < N + 1; ++i)
        cin >> vec[i];
    //可能か否かの判定
    vector<pair<int, int>> kanou(N + 1);
    kanou[N].first = vec[N];
    kanou[N].second = vec[N];
    for (int i = N - 1; i >= 0; --i)
    {
        kanou[i].first = kanou[i + 1].first / 2 + kanou[i + 1].first % 2 + vec[i];
        kanou[i].second = kanou[i + 1].second + vec[i];
    }
    // for (auto i : kanou)
    //     cout << i.first << " " << i.second << "\n";
    //グラフが作成できない場合
    if (kanou[0].first > 1)
    {
        cout << "-1\n";
        return 0;
    }

    vector<int> MAX(N + 1);
    MAX[0] = 1;
    for (int i = 1; i < N + 1; ++i)
        MAX[i] = min(kanou[i].second, 2 * (MAX[i - 1] - vec[i - 1]));
    cout << accumulate(MAX.begin(), MAX.end(), 0) << "\n";
}