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;

int main() {
  long long a=1,b=0,c=0,d=0,m=0,n=0,p=0,q=0;
  
  cin >> n;
  vector<long long> v(n+1,0),ok(n+1,0);
  c=pow(10,13);
  for(int i=0;i<n+1;i++)
  {
	  cin >> v.at(i);
	  ok.at(i)=a;
	  a -= v.at(i);
	  a *= 2;
    a=min(a,c);
  }
  a=0;
  for(int i=n;i>=0;i--)
  {
	 a+=v.at(i);
     if(ok.at(i)<a)
	 {   
         a=ok.at(i);
		 if(a-v.at(i)<=0)
		 {
			 cout << "-1";
			 return 0;
		 }
		 
	 }		 
    b+=a;
  }
  cout << b;
	  
  
}
