For two sequences S and T of length N consisting of 0 and 1, let us define f(S, T) as follows:

* Consider repeating the following operation on S so that S will be equal to T. f(S, T) is the minimum possible total cost of those operations.

* Change S_i (from 0 to 1 or vice versa). The cost of this operation is D \times C_i, where D is the number of integers j such that S_j \neq T_j (1 \leq j \leq N) just before this change.



There are 2^N \times (2^N - 1) pairs (S, T) of different sequences of length N consisting of 0 and 1. Compute the sum of f(S, T) over all of those pairs, modulo (10^9+7).

Constraints

* 1 \leq N \leq 2 \times 10^5
* 1 \leq C_i \leq 10^9
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
C_1 C_2 \cdots C_N


Output

Print the sum of f(S, T), modulo (10^9+7).

Examples

Input

1
1000000000


Output

999999993


Input

2
5 8


Output

124


Input

5
52 67 72 25 79


Output

269312
import java.util.Arrays;
import java.util.Scanner;

/**
 * Change a Little Bit
 * 
 * https://atcoder.jp/contests/abc150/tasks/abc150_e
 */
public class Main {

    public static void main(String[] args) throws Exception {
        try (Scanner sc = new Scanner(System.in)) {
            long MOD = 1000000007;

            int N = sc.nextInt();
            long[] C = new long[N];
            for (int i = 0; i < N; i++) {
                C[i] = sc.nextLong();
            }

            Arrays.sort(C);

            long P = pow(2, N - 1, MOD);

            long ans = 0;
            for (int i = 0; i < N; i++) {
                long p = P * (N - i + 1) % MOD;
                ans = (ans + C[i] * p) % MOD;
            }

            ans = (ans * P) % MOD;

            System.out.println(ans);
        }
    }

    static long pow(long A, long B, long MOD) {
        if (B == 0) {
            return 1;
        } else if (B % 2 == 0) {
            long C = pow(A, B / 2, MOD);
            return C * C % MOD;
        } else {
            long C = pow(A, B - 1, MOD);
            return A * C % MOD;
        }
    }

}
