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 Rust (1.42.0) (judge verdict: Wrong Answer)

use proconio::{fastout, input};
// use proconio::marker::*;

#[fastout]
fn main() {
   input! {
      n: usize,
      mut a: [usize; n+1]
   }
   if a[0] != 0 {
      if n == 0 && a[0] == 1 {
         println!("1");
      } else {
         println!("-1");
      }
      return;
   }

   let mut mn = vec![0usize; n + 1];
   let mut mx = vec![0usize; n + 1];
   mn[n] = a[n];
   mx[n] = a[n];
   for i in (0..n).rev() {
      mn[i] = a[i] + (mn[i + 1] - 1) / 2 + 1;
      mx[i] = mx[i + 1] + a[i];
      if mx[i] >> i != 0 {
         mx[i] = 1 << i;
      }
      if mx[i] < mn[i] {
         println!("-1");
         return;
      }
   }
   let mut now = 1;
   let mut ans = 0;
   for i in 0..n {
      ans += now;
      let m = (now-a[i]) * 2;
      if m < mn[i+1] {
         println!("-1");
         return;
      }
      now = std::cmp::min(mx[i+1], m);
   }
   ans += now;
   println!("{}", ans);
}