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.*;
import java.io.*;

public class Main {
    public static void main(String[] args) throws IOException{
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        int n = Integer.parseInt(f.readLine());
        long[] arr = new long[n+1];
        StringTokenizer st = new StringTokenizer(f.readLine());
        long sum = 0;
        for(int i = 0; i < n+1; i++){
            arr[i] = Long.parseLong(st.nextToken());
            sum+=arr[i];
        }
        long currLeaves = 1;
        long totalNodes = 1;
        boolean good = true;
        if(n != 0 && arr[0] != 0){
            out.println(-1);
            out.close();
            System.exit(0);
        }
        if(n == 0 && arr[0] == 1){
            out.println(1);
            out.close();
            System.exit(0);
        }else if(n == 0){
            out.println(-1);
            out.close();
            System.exit(0);
        }
        for(int i = 1; i <= n; i++){
            currLeaves = Math.min(2*currLeaves, sum);
            totalNodes+=currLeaves;
            if(arr[i] > currLeaves){
                good = false;
                break;
            }
            currLeaves -= arr[i];
            sum-= arr[i];
        }
        if(good){
            out.println(totalNodes);
        }else{
            out.println(-1);
        }





        out.close();
    }
}
