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.Scanner;

public class Main {

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = Integer.parseInt(sc.nextLine());
    long answer = 0;
    String[] input = sc.nextLine().split(" ");
    int[] nodeCount = new int[n + 1];
    long[] maxNodeCount = new long[n + 1];
    long[] minNodeCount = new long[n + 1];
    for(int i = 0; i <= n; i++){
      nodeCount[i] = Integer.parseInt(input[i]);
    }
    int idx = n;
    maxNodeCount[idx] = nodeCount[idx];
    minNodeCount[idx] = nodeCount[idx];
    idx--;
    while(idx >= 0){
      maxNodeCount[idx] = maxNodeCount[idx + 1] + nodeCount[idx];
      idx--;
    }
    minNodeCount[0] = 1;
    for (int i = 1; i <= n; i++){
      minNodeCount[i] = Math.min((minNodeCount[i -1] - nodeCount[i - 1]) * 2, maxNodeCount[i]);
      if(minNodeCount[i] < nodeCount[i]){
        System.out.print(-1);
        return;
      }
    }
    if(minNodeCount[n] != nodeCount[n]){
      System.out.print(-1);
      return;
    }
    for(long cnt : minNodeCount){
      answer += cnt;
    }
    System.out.print(answer);
  }
}