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"
using namespace std;
#define REP(i, n) for(ll i = 0;i < n;i++)
#define ll long long
#define MOD 1000000007LL
#define llMAX 9223372036854775807LL
#define llMIN -9223372036854775808LL
using vi = vector<ll>; // intの1次元の型に vi という別名をつける
using vvi = vector<vi>; // intの2次元の型に vvi という別名をつける
 
ll n,k,x,y;
 
int main(){
  ll ans=0;
  cin >> n ;
  vi a(n+1),b(n+1),c(n+1),d(n+1);
  REP(i,n+1) cin>>a[i];



  if (n==0){
    if (a[0]==1){
      cout<<1<<endl;
      return 0;
    }else{
      cout<<-1<<endl;
      return 0;
    }
  }
  


  if (a[0]!=0){
    cout<<-1<<endl;
    return 0;
  }

 
 

  d[n]=a[n];
  for(ll i=n-1;i>=0;i--){
    d[i]=d[i+1]+a[i];
  }

  ll nown=1;
  b[0]=1;
  c[0]=1;
  for(ll i=1;i<n+1;i++){
    nown=min(min(nown*2,d[i]),llMAX);
    b[i]=nown;
    nown-=a[i];
    if (nown<0){
      cout<<-1<<endl;
      return 0;
    }


    c[i]=nown;
  }

  

 
 
 
  int flag=0;
  for(ll i=0;i<n+1;i++){
    ans+=b[i];
    flag+=(b[i]<=0);
    //flag+=(c[i]<0);
  }
  if (flag!=0){
    cout<<-1<<endl;
    return 0;
  }
 
 
 
  cout<<ans<<endl;
  return 0;
}
 
/*
10
0 0 1 1 2 3 5 8 28 21 34
*/