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 <bits/stdc++.h>
#define rep(i,n) for(int i=0; i<n; i++)
#define PI 3.14159265359
#define INF 1000100100
#define MOD 1000000007
#define all(x) (x).begin(),(x).end()
typedef long long ll;
#define P pair<ll,ll>
#define PP pair<P,P>
using namespace std;

int main(){
    int n; cin >> n;
    vector<ll> a(n+1);
    vector<ll> wa(n+1);
    rep(i,n+1) {
        cin >> a[i];
    }
    wa[n]=a[n];
    rep(i,n){
        wa[n-i-1]=wa[n-i]+a[n-i-1];
    }
    ll ans=1;
    ll pa=1;
    if(a[0]!=0){
        if(n==0 && a[0]==1){
            cout << 1 << endl;
            return 0;
        }
        cout << -1 << endl;
        return 0;
    }
    rep(i,n){
        pa=min(wa[i+1],pa*2);
        //cout << i << ' ' << n-1 << endl;
        if(pa-a[i+1]<=0 && i!=n-1){
            cout << -1 << endl;
            return 0;
        }
        if(i==n-1 && pa-a[i+1]<0){
            cout << -1 << endl;
            return 0;
        }
        ans+=pa;
        pa-=a[i+1];
    }
    cout << ans << endl;
return 0;
}