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;
#define ll long long
int main(){
  ll N,answer=1,dp[100020];
  cin>>N;
  vector<ll> vec(N+1);
  vector<ll> sum(N+1);
  for(ll i=0;i<=N;i++){cin>>vec.at(i);}
  dp[0]=1;
  sum.at(N)=vec.at(N);
  for(ll i=N-1;i>=0;i--){
    sum.at(i)=sum.at(i+1)+vec.at(i);
  }
  for(ll i=1;i<=N;i++){
    dp[i]=min(2*(dp[i-1]-vec.at(i-1)),sum.at(i));
    if(dp[i]<=0){cout<<-1<<endl;return 0;}
    answer+=dp[i];
  }
  if(dp[N]!=vec.at(N)){cout<<-1<<endl;}
  else{cout<<answer<<endl;}
}