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.io.*;
import java.util.*;

public class Main {
    static PrintWriter out = new PrintWriter(System.out);
    static Scanner sc = new Scanner(System.in);
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    long mod = 998244353;
    long inv2 =1;
    public static void main(String[] args) throws IOException {
        Main main = new Main();
//        main.solve();
        out.println(main.solve());
        out.flush();
    }
    long solve() throws IOException {
        int n = sc.nextInt();
        long[] a = new long[n+1];
        for(int i=0; i<=n; i++) a[i] = sc.nextLong();
//        if(a[0]>0) return -1;
        long[] mini = new long[n+1], maxi = new long[n+1];
        mini[n] = a[n]; maxi[n] = a[n];
        for(int i=n-1; i>=0; i--){
            mini[i] = (mini[i+1]+1)/2;
            mini[i] += a[i]; maxi[i] = maxi[i+1]+a[i];
        }
        if(mini[0]!=1) return -1;
        long cur = 1, ans = 1;
        for(int i=1; i<=n; i++){
            cur = Math.min(cur*2, maxi[i]);
            ans += cur;
            cur -= a[i];
        }
        return ans;
    }
}