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>

using namespace std;

const int N = 1e5 + 7 , mod = 1e9 + 7 ;

unsigned long long n , m; 
string s ;
unsigned long long l[N] , r[N]; 
int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);
       // freopen("in.in" , "r" , stdin) ; 
        cin>>n ; 
        unsigned long long a[n+1] ; 
        unsigned long long ans = 0 , mx = 1 ;         
        for(int i = 0 ;i <=n;i++){
                unsigned long long t ; 
                cin>>t ;
                a[i] = t ;  
                if(t > mx){
                        return cout<<-1,0 ; 
                }
                r[i] = mx ; 
                mx-=t ; 
                mx*=2 ; 
        }
        
        unsigned long long all = a[n] ; 
        ans = a[n] ; 
        for(int i = n ;i >0 ; i--){
                unsigned long long rem =min(r[i-1] - a[i-1] ,  (1<<(i-1)) - a[i-1] )  ;
                unsigned long long tws = 0 , ons = 0 ; 
                if(all >= rem){
                        ons = min(rem , all) ; 
                        all-=ons ; 
                        tws = all ; 
                        ons-=tws ; 
                }
                else {
                        ons = all ; 
                }
                ans+= ons + tws + a[i-1];         
                all = ons + tws + a[i-1]  ; 
                
        }

        cout<< ans ;  
        return 0;
}