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<iostream>
#include <cstring>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
using namespace std;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
using ll = long long;
typedef pair<int,int> pint;
typedef pair<ll,int> pli;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;
const ll mod=1e9+7 ;



int main(){
  ll N; cin >> N; N++; ll a[N];
  rep(i, N){
    cin >> a[i];
  }
  ll tmp=0, ans=0;
  ll need[N] = {}; need[0] = 1;
  if(a[0]!=0){
    if(N==1&&a[0]==1){cout << 1 << endl; return 0;}
    else {cout << -1 << endl; return 0;}
  }
  if(N==1){ cout << -1 << endl; return 0;}
  rep(i, N-1){
    need[i+1] = (need[i]-a[i])*2;
    if(need[i+1]<=0){
      cout << -1 << endl; return 0;
    }
  }
  ll v[N];
  for(int i=N-1; i>=0; i--){
    tmp+=a[i];
    v[i] = min(need[i], tmp);
    tmp=v[i];
  }
  rep(i, N){
    ans+=v[i];
  }
  cout << ans << endl;
}