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.Arrays;
import java.util.Scanner;

public class Main {

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    long[] array = new long[n + 1];
    for(int i = 0; i <= n; i++){
      array[i] = sc.nextLong();
    }
    System.out.println(solve(array));
  }
  static long solve(long[] a){
    long[] nodeCount = new long[a.length];
    long prev = 0;
    for(int i = a.length - 1; i > 0; i--){
      nodeCount[i] = prev + a[i];
      prev = nodeCount[i];
    }
    nodeCount[0] = 1;

    for(int i = 1; i < a.length; i++){
      long tmp = nodeCount[i - 1] - a[i - 1];
      if(tmp <= 0){
        return -1;
      } else {
        nodeCount[i] = Math.min(nodeCount[i], tmp * 2);
      }
    }
    if(nodeCount[nodeCount.length - 1] != a[a.length - 1]){
      return -1;
    }
    return Arrays.stream(nodeCount).sum();
  }
}
