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: Accepted)

#include <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cctype>
#include <string>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <functional>
using namespace std;
typedef long long ll;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    ll n;
    cin>>n;
    
    ll total = 0ll;
    vector<ll> s(n+1,0ll);
    for(ll i =0; i<=n; i++){
        ll e;
        cin>>e;
        s[i]=e;
        total += e;
    }
    
    if(s[0]>1){
        cout<<-1<<endl;
        return 0;
    }
    
    if(s[0]==1 && total>1){
        cout<<-1<<endl;
        return 0;
    }
    
    if(s[0]==1 && total==1){
        cout<<1<<endl;
        return 0;
    }
    
    ll ans = 1ll;
    
    ll sum = 1;
    
    
    for(int i=1; i<=n; i++){
        ll m = min(2*sum, total);
        ans+=m;
        sum = m-s[i];
        if(sum<0){
            cout<<-1<<endl;
            return 0;
        }
        total -= s[i];
        
    }
    cout<<ans<<endl;
    
    return 0;
}