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;
int n;
int a[1000010];
long long ksm(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % 1000000007;
    a = a * a % 1000000007;
    b >>= 1;
  }
  return ret;
}
long long A = 1, ans;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    A = A * i % 1000000007;
  }
  sort(a + 1, a + 1 + n);
  int pos;
  for (int i = 1; i <= n; i = pos) {
    pos = i;
    while (a[i] == a[pos] && pos <= n) pos++;
    if (pos <= n)
      ans = (ans + A * ksm(n - i + 1, 1000000007 - 2) % 1000000007 * (pos - i) %
                       1000000007 * a[i] % 1000000007) %
            1000000007;
  }
  cout << ans << endl;
  return 0;
}
