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<string>
#include<algorithm>    
#include<cmath>
#include<map>
#include<vector>
#include<math.h>
#include<stdio.h>
#include<stack>
#include<queue>
#include<tuple>
#include<cassert>
#include<set>
//#include<bits/stdc++.h>
using ll = long long;
using namespace std;
const ll INF = 1000000000000000000;
const ll mod = 1000000007;
ll test[12345678], res[12345678];
signed main() {
	ll n, x = 0, cnt = 0, ans = 0;
	cin >> n;
	for (int h = 0; h <= n; h++) {
		cin >> test[h];

		if (h == 0) { res[h] = 1; }
		else { res[h] = res[h - 1] * 2 - test[h]; }

		if (h != n && res[h] <= 0) { x++; }
		if (h == n && res[h] < 0) { x++; }
		if (h == 0 && test[h] != 0) { x++; }
	}
	/*if (n == 0) {
		if (test[0] == 1) { cout << 1 << endl; }
		else { cout << -1 << endl; }
		return 0;
	}*/
	if (x != 0) { cout << -1 << endl; return 0; }

	cnt = test[n]; ans += cnt;

	for (int h = n - 1; h >= 0; h--) {
		if (cnt > res[h]) { cnt = res[h]; }
		cnt += test[h]; ans += cnt;
	}
	cout << ans << endl;
	return 0;
}