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++)
using namespace std;
using P = pair<int, int>;
typedef long long ll;

int main() {
  int n;
  scanf("%d", &n);
  vector<ll> a(n+1);
  rep(i, n + 1) cin >> a[i];
  vector<ll> Lower(n + 1);
  vector<ll> Upper(n + 1);
  Lower[n] = Upper[n] = a[n];
  for (int i = n - 1; i >= 0; i--){
    Lower[i] = a[i] + (Lower[i + 1] + 1) / 2;
    Upper[i] = a[i] + Upper[i + 1];
  }
  if(Lower[0] > 1){
    cout << -1 << endl;
    return 0;
  }
  ll cur = 1;
  ll ans = 0;
  rep(i, n+1){
    ans += cur;
    cur -= a[i];
    cur = min(cur * 2, Upper[i + 1]);
  }
  cout << ans << endl;
}