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;

using ll = long long;
using ld = long double;

using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vs = vector<string>;
using vld = vector<ld>;
using vvld = vector<vld>;

typedef pair<ll, ll> P;

#define bit(n) (1LL << (n))

//#define int long long

#define all(v) v.begin(), v.end()

#define rep(i, n) for (ll i = 0; i < n; i++)
#define REP(i, n) for (ll i = 1; i < n; i++)

#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define FORm(i, m) for (auto i = m.begin(); i != m.end(); i++)

template <class T>
inline void chmax(T& a, T b) {
  a = std::max(a, b);
}
template <class T>
inline void chmin(T& a, T b) {
  a = std::min(a, b);
}

#define mod (ll)(1e9 + 7)
const long long INF = 1LL << 60;

signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);

  ll n;
  cin >> n;
  vll a(n + 1);
  rep(i, n + 1) {
    cin >> a[i];
  }

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

  vll sum_leaf(n + 2);
  for (int i = n; i >= 0; i--) {
    sum_leaf[i] = sum_leaf[i + 1] + a[i];
  }

  vll num_node(n + 1);
  num_node[0] = 1;
  REP(i, n + 1) {
    num_node[i] = min(sum_leaf[i], (num_node[i - 1] - a[i - 1]) * 2);

    if (num_node[i] < a[i]) {
      cout << -1 << endl;
      return 0;
    }
  }

  cout << accumulate(all(num_node), 0ll) << endl;

  return 0;
}