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 <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
 
using namespace __gnu_pbds;
using namespace std;
 
typedef long long int ll;
typedef unsigned long long int sll;
typedef  double ld;
#define A 1000000007ll
#define D 1000000000ll
#define B 998244353ll
#define C 1000000000000000000ll
#define FAST ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define oset tree<ll, null_type,less_equal<ll>, rb_tree_tag,tree_order_statistics_node_update> 
#define pb push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define mxe(v) *max_element(v.begin(),v.end())
#define mne(v) *min_element(v.begin(),v.end())
#define bs binary_search
#define lb lower_bound
#define ub upper_bound
#define ve vector
#define br break
#define PI acos(-1)
#define subt cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n"

int main() 
{
    FAST;
    ll n;
    cin>>n;
    ll a[n+1],avp[n+1];
    for(int i=0;i<=n;i++)
    {
        cin>>a[i];
        avp[i]=0;
    }
    ll la=1;
    for(int i=0;i<=n;i++)
    {
        if(a[i]>la)
        {
            cout<<-1;
            return 0;
        }
        avp[i]=la-a[i];
        la=min(2*avp[i],A);
    }
    ll ca=a[n],ans=ca;
    for(int i=n-1;i>=0;i--)
    {
        if(2*avp[i]<ca)
        {
            cout<<-1;
            return 0;
        }
        if(ca<avp[i])
            ca=ca+a[i];
        else
            ca=avp[i]+a[i];
        ans+=ca;
    }
    cout<<ans;
    subt;
    return 0;
}