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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#ifdef ENABLE_DEBUG
#define dump(a) cerr<<#a<<"="<<a<<endl
#define dumparr(a,n) cerr<<#a<<"["<<n<<"]="<<a[n]<<endl
#else
#define dump(a) 
#define dumparr(a,n) 
#endif
#define FOR(i, a, b) for(ll i = (ll)a;i < (ll)b;i++)
#define For(i, a) FOR(i, 0, a)
#define REV(i, a, b) for(ll i = (ll)b-1LL;i >= (ll)a;i--)
#define Rev(i, a) REV(i, 0, a)
#define REP(a) For(i, a)
#define SIGN(a) (a==0?0:(a>0?1:-1))

typedef long long int ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef pair<ll, ll> pll;
typedef pair<ll,pll> ppll;
typedef vector<ll> vll;
typedef long double ld;
typedef pair<ld,ld> pdd;

const ll INF=(1LL<<60);
#if __cplusplus<201700L
ll gcd(ll a, ll b) {
  a=abs(a);
  b=abs(b);
  if(a==0)return b;
  if(b==0)return a;
  if(a < b) return gcd(b, a);
  ll r;
  while ((r=a%b)) {
    a = b;
    b = r;
  }
  return b;
}
#endif
template<class T>
bool chmax(T& a,const T& b){
  if(a<b){
    a=b;
    return true;
  }
  return false;
}
template<class T>
bool chmin(T& a,const T& b){
  if(a>b){
    a=b;
    return true;
  }
  return false;
}
template<class S,class T>
std::ostream& operator<<(std::ostream& os,pair<S,T> a){
  os << "(" << a.first << "," << a.second << ")";
  return os;
}
template<class T>
std::ostream& operator<<(std::ostream& os,vector<T> a){
  os << "[ ";
  REP(a.size()){
    os<< a[i] << " ";
  }
  os<< " ]";
  return os;
}


void solve(long long N, std::vector<long long> A){
  if(N==0){
    if(A[0]==1){
      cout<<1<<endl;
    }else{
      cout<<-1<<endl;
    }
    return;
  }
  if(A[0]>0){
    cout<<-1<<endl;
    return ;
  }
  ll cur=1,ans=0;
  REP(N+1){
    if(cur<A[i]){
      cout<<-1<<endl;
      return;
    }
    if(cur<INF){
      cur-=A[i];
      cur*=2;
    }
  }
  cur=0;
  Rev(i,N+1){
    cur+=A[i];
    if(i<50){
      ll empty=0;
      For(j,i){
        empty+=A[j]*(1<<(i-j));
      }
      cur=min(cur,(1LL<<i)-empty);
    }
    ans+=cur;
  }
  cout<<ans<<endl;
}

int main(){
  cout<<setprecision(1000);
    long long N;
    scanf("%lld",&N);
    std::vector<long long> A(N-0+1);
    for(int i = 0 ; i < N-0+1 ; i++){
        scanf("%lld",&A[i]);
    }
    solve(N, std::move(A));
    return 0;
}