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>
#include <math.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define PI 3.14159265359
using namespace std;
using ll = long long;
using P = pair<int, int>;

int main() {
  int n;
  cin >> n;
  vector<ll> a(n+1);
  vector<ll> sum(n+1);
  rep(i, n+1) cin >> a[i];
  sum[n] = a[n];
  for (int i=n; i>0; i--) {sum[i-1] = sum[i] + a[i-1];}
  //rep(i, n) cout << sum[i] << endl;

  ll ans = 1;
  ll par = 1;
  rep(i, n) {
    par = 2 * (par - a[i]);
    if (par < a[i+1]) {cout << -1 << endl; return 0;}
    par = min(par, sum[i+1]);
    ans += par;
  }

  cout << ans << endl;

  return 0;
}

/*
10
0 0 1 1 2 3 5 8 13 21 34
*/