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>
#define mp make_pair
#define pii pair<int,int>
#pragma GCC optimize ("O3")
#define pdd pair<double,double>
#define pll pair<ll,ll>
#define mst(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double eps = 1e-9;
const int maxn = 2e5+9;
const double pi=3.14159265358979323846264;
const int inf = 2e9;
const ll mod = 998244353;
ll op[maxn],pq[maxn];
int main()
{
    ios::sync_with_stdio(false);
    int n;
    cin>>n;
    for(int i=0;i<=n;++i) cin>>op[i];
    if(op[0]){
        cout<<-1;
        return 0;
    }
    for(int i=n-1;i>=1;--i){
        pq[i]=pq[i+1]+op[i+1];
    }
//    for(int i=1;i<=n;++i) cout<<pq[i]<<",";
//    cout<<endl;
    ll la=1,fg=1,ans=1;
    for(int i=1;i<=n;++i){
        if(op[i]>la*2){
            fg=0;
            break;
        }
        la=min(pq[i],la*2-op[i]);
        ans+=la+op[i];
//        cout<<la<<","<<endl;
    }
    if(!fg) cout<<-1;
    else cout<<ans;
    return 0;
}