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 M = 1e6 + 10;
long long n, t[M], maks, s, v, ans, mod = 1e9 + 7, sum;
long long sp(long long p, long long w) {
  if (w == 1) return p;
  if (w % 2) return (sp(p, w - 1) * p) % mod;
  long long f = sp(p, w / 2);
  return (f * f) % mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int a = 1; a <= n; a++) {
    cin >> t[a];
    maks = max(maks, t[a]);
  }
  s = 1;
  for (long long a = 1; a <= n; a++) s = (s * a) % mod;
  sort(t + 1, t + n + 1);
  for (int a = 1; a <= n; a++) {
    if (t[a] == maks) break;
    if (t[a] != t[a - 1]) v = a - 1;
    sum = s * sp(n - v, mod - 2);
    sum %= mod;
    sum *= t[a];
    ans += sum;
    ans %= mod;
  }
  cout << ans << '\n';
  return 0;
}
