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
import java.util.*;

class Main{
  public static void main(String[] args){
    Scanner sc=new Scanner(System.in);
    int n=sc.nextInt();
    long[] a=new long[n+1];
    long[] r=new long[n+1];
    boolean fl=true;
    for(int i=0;i<n+1;i++){
      a[i]=sc.nextLong();
      if(a[i]!=0){
        fl=false;
      }
    }
    if(fl){
      System.out.println(0);
      return;
    }
    for(int i=n-1;i>=0;i--){
      r[i]=r[i+1]+a[i+1];
    }
//    System.err.println(Arrays.toString(r));
    long ans=0;
    long cur=1;
    for(int i=0;i<n+1;i++){
      ans+=cur;
      if(cur<a[i]){
        System.out.println(-1);
        return;
      }
      cur-=a[i];
      cur=Math.min(cur*2,r[i]);
    }
    System.out.println(ans);
  }
}