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)

// -----------------------------------
// author   : MatsuTaku
// country  : Japan
// created  : 05/30/20 16:02:55
// -----------------------------------

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);

  int n; cin>>n;
  vector<ll> A(n+1); for (auto& a:A) cin>>a;
  auto solve = [&]() -> ll {
    ll leaves = accumulate(A.begin(), A.end(), 0ll);
    auto l = leaves;
    ll w = 1;
    ll ans = 0;
    for (int d = 0; d <= n; d++) {
      ans += w;
      if (w < A[d])
        return -1;
      w -= A[d];
      l -= A[d];
      if (w < 0)
        return -1;
      if (w*2 <= l) {
        w *= 2;
      } else {
        auto diff = l-w;
        if (diff == 0)
          continue;
        int k = 63;
        while (k >= 0 and ((1ull<<k) & diff) == 0)
          k--;
        if (k >= 0) {
          w += 1ull<<k;
        }
      }
    }
    return ans;
  };
  cout << solve() << endl;

  return 0;
}