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 <vector>

using namespace std;

int main() {
    long long N;
    cin >> N;

    vector<long long> A(N + 1);
    for (long long i = 0; i < N + 1; ++i) {
        long long Ai;
        cin >> Ai;
        A.at(i) = Ai;
    }

    long long available = 1;
    bool foundZero = false;

    vector<long long> rest(N + 1);

    for (long long i = 0; i < N + 1; ++i) {
        long long r = available - A.at(i);

        if (r < 0) {
            cout << -1 << endl;
            return 0;
        } else if (r == 0) {
            foundZero = true;
        } else if (foundZero && A.at(i) > 0) {
            cout << -1 << endl;
            return 0;
        }

        available = r * 2;
        rest.at(i) = r;
    }

    long long sum = A.at(N);
    long long children = 0;

    for (long long i = N - 1; i >= 0; --i) {
        long long branch = A.at(i + 1) + children < rest.at(i) ? A.at(i + 1) + children : rest.at(i);
        sum += branch + A.at(i);
        children = branch + A.at(i);
    }

    cout << sum << endl;
}