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>
#pragma GCC optimize("O3")
using namespace std;
const long long MOD = (long long)(1e9 + 7);
long long n, arr[1000005], fact1[1000005], fact2[1000005];
long long wow(int ind) { return fact1[n - ind - 1] * fact2[n - ind + 1] % MOD; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  sort(arr + 1, arr + 1 + n);
  fact1[0] = fact2[n + 1] = 1;
  for (int i = 1; i <= n; i++) {
    fact1[i] = (fact1[i - 1] * i) % MOD;
  }
  for (int i = n; i > 0; i--) {
    fact2[i] = (fact2[i + 1] * i) % MOD;
  }
  long long ok = 0;
  long long ko = 0;
  long long fuck = 0;
  for (int i = 1; i <= n; i++) {
    if (arr[i] == arr[n]) break;
    if (arr[i] == arr[i - 1])
      ok++;
    else {
      ko += ok;
      ok = 1;
    }
    fuck = (fuck + (wow(ko) * arr[i]) % MOD) % MOD;
  }
  cout << fuck << '\n';
}
