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 <cstdio>
#include <cmath>
using namespace std;

const int N=100005;

int n,now,ans,flag;
int a[N];
double sum[N];

int power(int a,int b) {
	int r=1,base=a;
	while(b) {
		if(b&1) r*=base;
		base*=base;
		b>>=1;
	}
	return r;
}

int main() {
	cin>>n,n++;
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	for(int i=2;i<=n;i++) sum[i]=sum[i-1]+(double)a[i]/(double)power(2,i-1);
	ans+=a[n];
	for(int i=n-1;i>=1;i--) {
		if((int)(ceil(sum[i-1]+(double)(a[i+1]+a[i])/(double)power(2,i-1)))<=1)
			ans+=a[i+1]+a[i],a[i]+=a[i+1],flag=1;
		else
			ans+=a[i]+(int)(ceil((double)a[i+1]/2)),a[i]+=(int)(ceil((double)a[i+1]/2));
	}
	if(flag) cout<<ans;
	else cout<<-1;
	return 0;
}