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 int
#define MOD 1000000007
#define INF 1e18

using namespace std;

int main(void){
  ll n;
  cin >> n;
  
  vector<ll> a(n + 1);
  for (int i = 0; i < n + 1; i++){
    cin >> a[i];
  }

  vector<ll> sum(n + 2, 0);
  for (int i = n; i >= 0; i--){
    sum[i] = sum[i + 1] + a[i];
  }


  ll ans = 0, can_use = 1, ok = 1;
  if (n == 0){
    if (a[0] == 1)
      ans = 1;
    else
      ok = 0;
  }
  else if (n == 1){
    if (a[0] == 0 && a[1] == 1){
      ans = 2;
    }
    else if (a[0] == 0 && a[1] == 2){
      ans = 3;
    }
    else 
      ok = 0;
  }
  else {
    for (int i = 0; i < n + 1; i++){
      // n = 0, n = 1はコーナーケース
      if (i == 0){
        if (a[i] == 0){
          can_use = 1;
          ans++;
        }
        else {
          ok = 0;
          break;
        }
      }
      else if (i == n){
        if (a[i] <= 2*can_use)
          ans += a[i];
        else 
          ok = 0;
      }
      else {
        if (can_use <= 1e15){
          can_use = 2*can_use - a[i];
          ans += a[i] + min(can_use, sum[i + 1]);// 使う分;
        }
        else {
          ans += a[i] + sum[i + 1];
        }
      }


      if (can_use <= 0){
        ok = 0;
        break;
      }
    }
  }

  if (ok){
    cout << ans << endl;
  }
  else 
    cout << -1 << endl;

  return 0;
}