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;

int64_t MAX = 1000000000;

int main()
{
	int n;

	cin >> n;
	vector<int64_t> a(n + 1);
	vector<int64_t> maxi(n + 1, 1);
	bool flag = true;
	for (int i = 0; i <= n; i++) {
		int64_t a_in;
		cin >> a_in;
		a.at(i) = a_in;
		if (n != 0 && a.at(0) != 0) {
			flag = false;
			break;
		}
		if (i != n) {
			maxi.at(i + 1) = min((maxi.at(i) - a.at(i)) * 2, MAX);
			if (maxi.at(i + 1) <= MAX && a.at(i + 1) > maxi.at(i + 1)) {
				flag = false;
				break;
			}
		}
	}

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

	vector<int64_t> gyaku(n + 1);
	gyaku.at(n) = a.at(n);
	for (int i = n - 1; i >= 0; i--) {
		gyaku.at(i) = min(gyaku.at(i + 1) + a.at(i), MAX);
	}

	int64_t ans = 0;
	for (int i = 0; i <= n; i++) {
		ans += min(maxi.at(i), gyaku.at(i));
	}

	cout << ans << endl;
}