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 <cstdio>
#include <iostream>
#include <cstring>

int a[10100000];

typedef long long ll;

int main()
{
	int n; scanf("%d", &n);
	for(int i = 0; i <= n; i++) scanf("%d", &a[i]);
	if(a[0] != 0) return 0 * puts("-1");
	ll c = 1, ans = n + 1; a[n]--;
	int j = n;
	for(int i = 1; i <= n; i++)
	{
		if(a[i] > c) return 0 * puts("-1");
		ll ex = c - a[i]; ans += a[i]; a[i] = 0;
		c += ex;
		while(j >= 1 && ex >= a[j])
		{
			ans += 1ll * a[j] * (j - i + 1);
			ex -= a[j]; a[j] = 0; j--;
		}
		if(!j) return 0 * printf("%lld\n", ans);
		ans += 1ll * ex * (j - i + 1); a[j] -= ex;
	}
	printf("%lld\n", ans);
}