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>
using namespace std;

#define fio ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define all(a) a.begin(),a.end()
#define for0(i, n) for (int i = 0; i < n; i++)
#define for1(i, n) for (int i = 1; i <= n; i++)
#define loop(i,a,b) for (int i = a; i < b; i++)
#define bloop(i,a,b) for (int i = a ; i >= b;i--)
#define tc(t) int t; cin >> t; while (t--)
#define ll long long
#define pb push_back
#define sz(v) v.size()
#define pln(a)  cout<<a;
#define rr return 0;
#define rr1 return 1;
#define vr return;
#define int long long 

void solve()
{
  int n;
  cin>>n;
  vector<int>v(n+1);
  for0(i, n+1)  cin>>v[i];
  
  vector<int>mx(n+1);
  mx = v;
  bloop(i, n-1, 0) {
    mx[i] += mx[i+1];
  }
  
  
  int ans = 0, vi = 1;
  for0(i, n+1) {
    if(i != 0)
      vi = min(2*vi, mx[i]);
    ans += vi;
    vi -= v[i];
    if(vi < 0) {
      cout<<"-1";
      vr;
    }
  }
  
  cout<<ans;
}

signed main() {

  fio;
  
  
  //tc(t) {
    solve();
    pln("\n");
  //}
  
  rr;
}