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: Time Limit Exceeded)

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp>
#define ll long long int
#define pb push_back
#define fi first
#define se second
#define all(x) x.begin(),x.end()
#define mem(x,y) memset(x,y,sizeof(x))//memset only use to initialize 0 or -1 no other number.
#define pii pair<int,int>
#define pll pair<ll,ll>
#define INF 1e9
#define INFL 1e18;
#define mod 1000000007
//#define mod 998244353
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
using namespace __gnu_pbds; 
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> os;
typedef tree<pii,null_type,less<pii>,rb_tree_tag,tree_order_statistics_node_update> os_pair;

ll power(ll x,ll n){ll res =1;while(n>0){if(n%2==1){res=res*x;}x=x*x;n=n/2;}return res;}
ll powm(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}

//cout<< fixed << setprecision(10)

int main(){

    fast;
    int T;
    T=1;
    //cin>>T;
    while(T--){
    	ll n,i;
    	cin>>n;
    	ll a[n+1];
    	for(i=0;i<n+1;i++)
    	{
    		cin>>a[i];
		}
		ll b[n+1];
		b[n]=a[n];
		for(i=n-1;i>=0;i--)
		{
			b[i]=b[i+1]+a[i];
		}
		ll c[n+1];
		c[0]=1;
		for(i=1;i<n+1;i--)
		{
			c[i]=min(b[i],2*(c[i-1]-a[i]));
		}
		int x=1;
		ll ans=0;
		for(i=0;i<n+1;i++)
		{
			if(c[i]<a[i])
			{
				x=0;
				
			}
			ans+=c[i];
		}
		if(x==0)
		{
			cout<<-1<<"\n";
		}
		else
		{
			cout<<ans;
		}

    }

}