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;
#define rep(i, n) for (int i=0; i<(int)(n); i++)
#define all(x) x.begin(), x.end()
#define mod 1000000007
typedef long long ll;

int main(){
  ll n;
  cin >> n;
  vector<ll> vec(n+1);
  for(ll i=0; i<=n; i++) cin >> vec[i];
  ll res = 0;
  ll current = 0;
  vector<ll> mx(n+1);
  if(n == 0 && vec[0] != 1){
    cout << -1 << endl;
    return 0;
  }
  if(n != 0 && vec[0] != 0){
    cout << -1 << endl;
    return 0;
  }
  for(ll i=0; i<=n; i++){
    if(i == 0) mx[i] = (ll)1;
    else{
      if((mx[i-1]-vec[i-1])*2 < vec[i]){
        cout << -1 << endl;
        return 0;
      }
      mx[i] = (mx[i-1]-vec[i-1])*(ll)2;
    }
  }
  for(ll i=n; i>=0; i--){
    if(i == n){
      current += vec[i];
      res += current;
    }else{
      current = min(vec[i]+current, mx[i]);
      res += current;
    }
  }
  cout << res << endl;
  return 0;
}