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 ll long long
#define rep(i,n) for(int i=0;i<n;i++)
const ll MOD=1000000007;
const long double PI=3.14159265358979;
const ll MAX=0;

int main() {
  ll N,c=1;
  cin>>N;
  ll A[N+1];
  for(int i=0;i<N+1;i++){
    cin>>A[i];
  }
  ll a=1;
  for(int i=0;i<N+1;i++){
    a-=A[i];
    if(i!=N&&a<=0){
      c=0;break;
    }
    a*=2;
    if(a>1000000000){
      a=1000000000;
    }
    //cout<<a<<endl;
  }
  ll ans=0;
  for(int i=0;i<N+1;i++){
    ans+=(i+1)*A[i];
  }
  ll A2[N+1];
  for(int i=N;i>=0;i--){
    if(i==N){
      A2[i]=A[i];
    }
    else{
      A2[i]=A2[i+1]+A[i];
    }
  }
  ll a2=1;
  for(int i=0;i<N+1;i++){
    if(a2<A2[i]){
      ans=ans-A2[i]+a2;
    }
    a2-=A[i];
    a2*=2;
    if(a2>1000000000){
      a2=1000000000;
    }
   // cout<<a2<<endl;
  }
  if(c){
    cout<<ans;
  }
  else{
    cout<<-1;
  }
}