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 int long long
template<typename T>
void fin(T a){
  cout<<a<<endl;
  exit(0);
}

signed main(){
  int n;cin>>n;
  if(n==0){
    int a;cin>>a;if(a==1)fin(1);fin(-1);
  }
  int a;cin>>a;if(a)fin(-1);
  vector<int> v(n);
  for(int i=0;i<n;i++)cin>>v[i];
  vector<int> w(n+1,0);
  for(int i=0;i<n;i++)w[i+1]=w[i]+v[n-1-i];
  int now=1,ans=1;
  for(int i=0;i<n;i++){
    if(now*2<v[i])fin(-1);
    int A=w[n-i-1];
    int B=min(now*2,v[i]+A);
    ans+=B;
    now=B-v[i];
  }
  fin(ans);
}
