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.
#include <bits/stdc++.h>
using namespace std;
long long n, ans, a[1000001], inv[1000001];
int main() {
  scanf("%I64d", &n);
  inv[1] = 1;
  for (long long i = 2; i <= n; i++)
    inv[i] = (1000000007 - 1000000007 / i) * inv[1000000007 % i] % 1000000007;
  for (long long i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  sort(a + 1, a + n + 1);
  for (long long i = 1, j = 1; i <= n; i = j) {
    while (a[j] == a[i] && j <= n) j++;
    if (j <= n)
      ans += (j - i) * a[i] % 1000000007 * inv[n - i + 1] % 1000000007;
  }
  for (long long i = 1; i <= n; i++) ans *= i, ans %= 1000000007;
  printf("%I64d\n", ans);
  return 0;
}
