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>
#include <algorithm>

#define INF 3223372036854775807LL

using namespace std;

int main()
{
	int n;
	cin >> n;
	vector<long long> a(n + 1);
	for (int i = 0; i < n + 1; i++) {
		cin >> a[i];
	}
		 
	if (n == 0) {
		cout << (a[0] == 1 ? -1 : 1) << endl;
		return 0;
	} else {
		if (a[0] != 0) {
			cout << 1 << endl;
			return 0;
		}
	}

	vector<long long> p(n + 1);
	p[0] = 1;
	for (int i = 1; i < n + 1; i++) {
		if (a[i] > p[i - 1] * 2) {
			cout << -1 << endl;
			return 0;
		} else {
			p[i] = p[i - 1] * 2 - a[i];
		}
	}
	
	vector<long long> pw(n + 1);
	pw[0] = 1;
	for (int i = 1; i <= n; i++) {
		pw[i] = min(INF, pw[i - 1] * 2);
	}

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

	vector<long long> ans(n + 1);
	ans[0] = 1;
	for (int i = 1; i <= n; i++) {
		ans[i] = min(ans[i - 1] * 2 - a[i], node[i]);
	}

	long long sum = 0;
	for (int i = 0; i < n; i++) {
		sum += a[i];
		sum += ans[i];
	}
	cout << sum + a[n] << endl;

	return 0;
}