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;
typedef long long int LL;
typedef pair<int,int> P;
typedef pair<LL,LL> PL;

const int MOD=1000000007;
int INF=100100100;
LL a[100010];
LL s[100010];
int main(){
  LL n;cin>>n;
  for(LL i=0;i<=n;i++)cin>>a[i];
  LL sum=0;
  for(LL i=n;i>=0;i--){
    sum+=a[i];
    s[i]=sum;
  }
  LL ans=0;
  LL nokori=1;
  for(LL i=0;i<=n;i++){
    //cout<<nokori<<" "<<s[i]<<endl;
    if(nokori<a[i]){
      cout<<-1<<endl;
      return 0;
    }
    ans+=min(nokori,s[i]);
    nokori-=a[i];
    nokori=min(2*nokori,s[i]);
   // cout<<ans<<endl;
  }
  cout<<ans<<endl;
  return 0;
}