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;
const int N = 1000000 + 5;
const int Mod = 1000000000 + 7;
long long inv[N];
long long a[N];
int n;
int main() {
  ;
  ;
  inv[1] = 1;
  inv[0] = 1;
  for (int i = 2; i < N; i++) {
    inv[i] = (Mod - Mod / i) * inv[Mod % i] % Mod;
  }
  while (~scanf("%d", &n)) {
    for (int i = 0; i < n; i++) scanf("%I64d", a + i);
    long long res = 0;
    sort(a, a + n);
    for (int i = 0; i < n;) {
      int j = i;
      while (i < n && a[i] == a[j]) i++;
      if (i != n) res += (((a[j] * inv[n - j]) % Mod) * (i - j)) % Mod;
      res %= Mod;
    }
    for (int i = 1; i <= n; i++) res = (res * i) % Mod;
    cout << res << endl;
  }
}
