You are given an array a of length n. We define fa the following way:

  * Initially fa = 0, M = 1; 
  * for every 2 ≤ i ≤ n if aM < ai then we set fa = fa + aM and then set M = i. 



Calculate the sum of fa over all n! permutations of the array a modulo 109 + 7.

Note: two elements are considered different if their indices differ, so for every array a there are exactly n! permutations.

Input

The first line contains integer n (1 ≤ n ≤ 1 000 000) — the size of array a.

Second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print the only integer, the sum of fa over all n! permutations of the array a modulo 109 + 7.

Examples

Input

2
1 3


Output

1

Input

3
1 1 2


Output

4

Note

For the second example all the permutations are:

  * p = [1, 2, 3] : fa is equal to 1; 
  * p = [1, 3, 2] : fa is equal to 1; 
  * p = [2, 1, 3] : fa is equal to 1; 
  * p = [2, 3, 1] : fa is equal to 1; 
  * p = [3, 1, 2] : fa is equal to 0; 
  * p = [3, 2, 1] : fa is equal to 0. 



Where p is the array of the indices of initial array a. The sum of fa is equal to 4.
// discussed with rainboy
import java.io.*;
import java.util.*;

public class CF938E {
	static final int MD = 1000000007;
	static long power(long a, int k) {
		if (k == 0)
			return 1;
		long p = power(a, k / 2);
		p = p * p % MD;
		if (k % 2 == 1)
			p = p * a % MD;
		return p;
	}
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		StringTokenizer st = new StringTokenizer(br.readLine());
		Integer[] aa = new Integer[n];
		for (int i = 0; i < n; i++)
			aa[i] = Integer.parseInt(st.nextToken());
		Arrays.sort(aa);
		long ans = 0;
		for (int i = 0, j; i < n; i = j) {
			int a = aa[i];
			for (j = i + 1; j < n && aa[j] == a; j++)
				;
			if (j < n)
				ans = (ans + power(n - i, MD - 2) * a % MD * (j - i)) % MD;
		}
		for (int i = 1; i <= n; i++)
			ans = ans * i % MD;
		System.out.println(ans);
	}
}
