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.

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.

Examples

Input

3
0 1 1 2


Output

7


Input

4
0 0 1 0 2


Output

10


Input

2
0 3 1


Output

-1


Input

1
1 1


Output

-1


Input

10
0 0 1 1 2 3 5 8 13 21 34


Output

264
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
constexpr ll mod=1e9+7;

int main()
{
  ll n;
  cin>>n;
  vector<ll> a(n+1);
  for(auto&& e: a) 
  {
    cin>>e;
  }

  vector<ll> nodes(n+1);
  nodes.back()=a.back();
  for (int i = n-1; i >= 0; --i)
  {
    nodes[i]=nodes[i+1]+a[i];
  }

  vector<ll> ans=nodes;
  ans[0]=1;
  for (int i = 1; i < nodes.size(); ++i)
  {
    if (ans[i-1]-a[i-1]<0) {
      cout<<-1<<endl;
      return 0;
    }
    ans[i]=min(nodes[i],2*(ans[i-1]-a[i-1]));
  }
  if (ans.back()!=a.back()) cout<<-1<<endl;
  else cout<<accumulate(ans.begin(), ans.end(),0LL)<<endl;
}