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;
typedef long long ll;

#define MOD (long long)(1e9+7)
#define rep(i,n) for(ll i = 0; i < (n); i++)
#define rep1(i,n) for(ll i = 1; i <= (n); i++)

void solve()
{
  ll n; cin >> n;

  vector<ll> a(n + 1); rep(i, n + 1) cin >> a[i];
  ll ans = 0;
  vector<ll> b(n + 1);
  if(n > 0 && a[0] != 0 || a[0] > 1) {
    cout << -1 << endl;
    return;
  }
  b[0] = 1;
  rep1(i, n) {
    b[i] = b[i -1] * 2 - a[i];
  }

  vector<ll> c(n + 1);
  c[n] = 0;

  for(ll i = n - 1; i >= 0; i --) {
    if(c[i + 1] + a[i + 1] > b[i] * 2) {
      cout << -1 << endl;
      return ;
    }
    if(c[i + 1] + a[i + 1] > b[i]) {
      c[i] = b[i];
    } else {
      c[i] = c[i + 1] + a[i + 1];
    }
  }

  rep(i, n + 1) {
    cerr << a[i] << " " << b[i] << " " << c[i] << endl;
    ans += c[i] + a[i];
  }
  cout << ans << endl;

}


int main(void)
{
  // ll t; cin >> t; REP(i, t)
  solve();
}