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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <utility>
#include <string>

using namespace std;

const long long maxi = 1e18;

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

	vector<long long> a(n + 1, 0);
	for (int i = 0; i <= n; i++) {
		cin >> a[i];
	}

	vector<long long> capacity(n + 1, 0);
	for (int i = 0; i <= n; i++) {
		long long tmp = (i == 0 ? 1 : 2LL * capacity[i-1]);
		if (i < n) {
			tmp -= a[i];
		}

		tmp = min(tmp, maxi);

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

	long long ans = 0;
	long long n_node = 0;
	for (int i = n; i >= 0; i--) {

		if (n_node > 2LL * capacity[i]) {
			ans = -1;
			break;
		}

		n_node = min(n_node, capacity[i]);
		n_node += a[i];

		ans += n_node;
	}

	cout << ans << endl;
	return 0;
}