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: Runtime Error)

#include "iostream"
#include "vector"
#include "cmath"

using namespace std;
typedef long long ll;

int main()
{
  ll n;
  cin >> n;

  vector<ll> as(n + 1, 0);
  for (ll i = 0; i <= n; i++)
    cin >> as[i];

  vector<ll> sums(n + 1, 0);
  // cout << "sums: ";
  for (ll i = n; i >= 0; i--)
  {
    if (i == n)
      sums[i] = as[i];
    else
      sums[i] = as[i] + sums[i + 1];

    // cout << sums[i];
  }
  // cout << endl;

  // cout << "as: ";
  // for (auto i = 0; i <= n; i++)
  // {
  //   cout << as[i] << " ";
  // }
  // cout << endl;

  if (n == 0)
  {
    if (as[0] != 1)
    {
      // cout << -1 << endl;
      return -1;
    }
    else
    {
      return 1;
    }
  }

  // 以下は深さ1以上.
  ll cnt = 1;
  ll last_has_child = 1;
  for (ll i = 1; i <= n; i++)
  {
    ll cap = min(2 * last_has_child, sums[i]);
    if (as[i] > cap)
    {
      cout << -1 << endl;
      return 0;
    }

    ll has_child = min(cap - as[i], sums[i]);

    // printf("i: %lld, child: %lld, leaf: %lld\n", i, has_child, as[i]);
    cnt += has_child + as[i];
    last_has_child = has_child;
  }

  cout << cnt << endl;

  return 0;
}