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<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pll;
#define rep(i,n) for(int i=0;i<n;i++)
#define all(x) x.begin(),x.end()
#define chmin(a,b) a=min(a,b)
#define chmax(a,b) a=max(a,b)
#define N 100010
ll n,a[N],b[N];
int main(){
	cin>>n;
	for(int i=0;i<=n;i++)cin>>a[i];
	if(a[0]>0){
		cout<<-1<<endl;
		return 0;
	}
	b[0]=1;
	for(int i=1;i<=n;i++){
		b[i]=2*b[i-1]-a[i];
		if(b[i]<0){
			cout<<-1<<endl;
			return 0;
		}
	}
	b[n]=0;
	for(int i=n-1;i>=0;i--){
		chmin(b[i],b[i+1]+a[i+1]);
	}
	ll ans=0;
	for(int i=0;i<=n;i++){
		ans+=a[i]+b[i];
	}
	cout<<ans<<endl;
}