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>
#define ll long long
using namespace std;


const long long MAXVAL = (long long) 1e18 + 1;
const long long MOD = 1000000007;
const int INF = 1000000000 + 1;

long long poww(int x, long long pow) {
  if (pow == 0ll ) return 1;
  if (pow == 1ll) return x % MOD;
  long long ans = poww(x, pow >> 1) % MOD;
  ans = (ans * ans) % MOD;
  if ((pow & 1) == 1) {
    ans = (ans * x ) % MOD;
  }
  return ans % MOD;
}

ll gcd(ll x, ll y) {
  if (x == 0) return y;
  while (y) {
    x %= y;
    swap(x, y);
  }
  return x;
}

double dfs(int v, vector<bool>& visited, vector<vector<int>>& gr) {
  visited[v] = true;
  double res = 0.;
  int num = 0;
  for (int next : gr[v]) {
    if (!visited[next]) {
      num++;
      double resv = dfs(next, visited, gr);
      res += resv;
    }
  }
  if (num == 0) {
    return 0;
  }
  res /= num;
  return res + 1;
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int q = 1;
  //cin >> q;
  cout << fixed << setprecision(6);

  while (q-- >0) {
    int n;
    cin >> n;
    int a[n + 1];
    ll maxnonleaf[n + 1];
    ll cur = 1ll;
    const ll INF = 100000000000000000ll;
    bool fail = false;
    for (int i = -1; i < n; ++i) {
      cin >> a[i + 1];
      if (cur >= INF) maxnonleaf[i + 1] = INF;
      else {
        maxnonleaf[i + 1] = (cur - a[i + 1]);
        if (maxnonleaf[i + 1] < 0) fail = true;
        cur = min(INF, 2* ( cur - a[i + 1]));
      }
    }
    if (fail) {
      cout << -1 << endl;
      continue;
    }
    ll res = 0ll;
    cur = 0;
    for (int i = n; i >= 0; --i) {
      cur = min(cur, maxnonleaf[i]) + a[i];
      res += cur;
    }
    cout << res << endl;
  }
  return 0;
}