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: Accepted)

#include <bits/stdc++.h>
using namespace std;

bool chmin(long long &a, long long b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}

int main() {
	int n;
	cin >> n;
	vector<int> a(n + 1);
	for (int i = 0; i < n + 1; i++) {
		cin >> a[i];
	}
	
	vector<long long> b(n + 1, a[n]);
	for (int i = n; i--> 0;) {
		b[i] = b[i + 1] + a[i];
	}
	
	long long v = 1;
	for (int i = 0; i < n + 1; i++) {
		if (chmin(b[i], v)) {
			v = (v - a[i]) * 2LL;
		}
	}
	
	long long ans = 0;
	bool ok = true;
	for (int i = 0; i < n + 1; i++) {
		ok &= (b[i] >= a[i]);
		ans += b[i];
	}
	
	if (not ok) ans = -1;
	
	cout << ans << '\n';
	
	return 0;
}