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;

int N;
int A[100001];

int main() {

    cin >> N;
    for (int i = 0; i < N + 1; i++) cin >> A[i];
    //今の深さの点の数
    int now = 1;
    //int mae = 0;
    long long ans = 0;
    now = 1;
    //末端の数
    long long sum = 0;
    //分岐した回数
    long long route = 1;
    
    for (int i = 0; i < N + 1; i++)
    {
        sum += A[i];
    }
    
    //cout << sum << endl;
    for (int i = 0; i < N+1; i++)
    {
        //cout << now <<route<< endl;
        ans += now;
        now -= A[i];
        //できるだけ早く分岐しておく
        int a = now;
        if (route < sum)
        {
            for (int j = 0; j < a; j++)
            {
                if (route < sum)
                {
                    now++;
                    route++;
                }
            }
        }
        if (i != N && now <= 0)
        {
            cout << -1 << endl;
            return 0;
        }
        else if (i == N && now < 0)
        {
            cout << -1 << endl;
            return 0;
        }
    }
    if (now != 0)
    {
        cout << -1 << endl;
        return 0;
    }
    /*
    long long del = 0;
    for (int i = 0; i < N + 1; i++)
    {
        //cout << now <<route<< endl;
        //ans += now;
        del += sum - route;
        now -= A[i];
        //できるだけ早く分岐しておく
        int a = now;
        
        for (int j = 0; j < a; j++)
        {
            if (route < sum)
            {
                now++;
                route++;
                
                if (route == sum) break;
            }
        }
        if (i != N && now <= 0)
        {
            cout << -1 << endl;
            return 0;
        }
        else if (i == N && now < 0)
        {
            cout << -1 << endl;
            return 0;
        }
        if (route == sum) break;
    }
    if (now != 0)
    {
        cout << -1 << endl;
        return 0;
    }
    for (int i = 0; i < N + 1; i++)
    {
        ans += (i + 1) * A[i];
    }
    ans -= del;
    */
    cout << ans << endl;
    return 0;
}