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;

typedef long long int LL;

#define st first
#define nd second
#define PLL pair <LL, LL>
#define PII pair <int, int>

const int N = 1e6 + 7;
const LL MX = 1e9 + 7;
const LL INF = 1e18 + 9LL;

int n;
LL in[N];
LL maxi[N];

int main(){
	scanf("%d", &n);
	for(int i = 0; i <= n; ++i)
		scanf("%lld", &in[i]);
	
	if(in[0] >= 1){
		if(n == 0)
			puts(in[0] == 1 ? "1" : "-1");
		else
			puts("-1");
		exit(0);
	}
	
	maxi[0] = 1;
	for(int i = 1; i <= n; ++i){
		maxi[i] = min(INF, 2 * (maxi[i - 1] - in[i - 1]));

		if(maxi[i] < in[i]){
			puts("-1");
			exit(0);
		}
	}

	LL ans = 0;
	for(int i = n; i >= 0; --i){
		ans += min(in[i], maxi[i]);
		if(i)
			in[i - 1] += in[i];
	}

	printf("%lld\n", ans);
	return 0;
}