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 MX = 1000005, mod = 1e9 + 7;
int n;
long long a[MX], fac[MX], inv[MX], mn = 0, acu, act, res = 0;
void precal() {
  fac[0] = 1;
  inv[0] = inv[1] = 1;
  for (int i = 1; i < MX; i++) fac[i] = (fac[i - 1] * i) % mod;
  for (int i = 2; i < MX; i++) inv[i] = inv[mod % i] * (mod - mod / i) % mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  precal();
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  int i = 0;
  while (i < n) {
    acu = 0;
    act = a[i];
    while (i < n && a[i] == act) {
      i++;
      acu++;
    }
    if (i < n) {
      res += act * acu % mod * fac[n] % mod * inv[n - mn] % mod;
      res %= mod;
    }
    mn += acu;
  }
  cout << ((((res) % (mod)) + (mod)) % (mod)) << '\n';
  return 0;
}
