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>
#define REP(i, n) for (int i = 0; i < (n); i++)
#define MOD 1000000007
#define INF 1e9
#define All(x) (x).begin(), (x).end()
typedef long long ll;
using namespace std;

int main() {
  ll N;
  cin >> N;
  N++;
  ll ans = 1;
  vector<ll> a(N), sum(N);
  REP(i, N) cin >> a[i];
  sum[N - 1] = a[N - 1];
  for (int i = N - 2; i >= 0; i--) {
    sum[i] = sum[i + 1] + a[i];
  }
  ll bef = 1 - a[0];
  if(bef==0&&N==1){
    cout << -1 << endl;
    return 0;
  }
  for (int i = 1; i < N; i++) {
    ll con = bef * 2 - a[i];
    if((con<=0&&i!=N-1)||con<0||(bef>a[i]&&i==N-1)){
      cout << -1 << endl;
      return 0;
    }
    ans += (min(sum[i] - a[i], con) + a[i]);
    bef = min(sum[i] - a[i], con);
  }
  
  cout << ans << endl;

  return 0;
}