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>

using namespace std;
using llong = long long;
using ldbl = long double;
using lpair = pair<llong, llong>;

#define ALL(x) x.begin(), x.end()

constexpr llong mod = 1e9+7;
constexpr llong inf = mod * mod;

int main() {
  llong N;
  vector<llong> A, B;
  llong ans = 0;

  cin >> N;
  A.resize(N + 1);
  B.resize(N + 1);
  B[0] = 1;
  llong sho = 0;
  for (int i = 0; i <= N; ++i) {
    if (i > 0) {
      B[i] = B[i - 1] * 2;
      sho *= 2;
      while (B[i] >= 2 * inf) {
        ++sho;
        B[i] -= inf;
      }
    }
    cin >> A[i];
    B[i] -= A[i];
    if (B[i] <= 0 && sho > 0) {
      B[i] += inf;
      --sho;
    }

    if (B[i] <= 0 && i < N) {
      ans = -1;
      break;
    }
  }

  if (ans == -1) {
    cout << ans << endl;
    return 0;
  }

  llong child = 0;
  ans = 0;
  for (int i = N; i >= 0; --i) {
    child = min(child, B[i]) + A[i];
    ans += child;
  }

  cout << ans << endl;

  return 0;
}