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 maxn = 1e6 + 10;
const long long mod = 1e9 + 7;
long long a[maxn], fac1[maxn], fac2[maxn];
int n;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%lld", &a[i]);
  sort(a, a + n);
  fac1[0] = 1;
  fac2[n + 1] = 1;
  fac2[n] = n;
  for (int i = 1; i <= n; ++i) fac1[i] = fac1[i - 1] * i % mod;
  for (int i = n - 1; i >= 1; --i) fac2[i] = fac2[i + 1] * i % mod;
  long long ans = 0;
  for (int i = 0, j = 1; i < n && a[i] != a[n - 1]; i = j) {
    for (j = i; a[i] == a[j]; j++)
      ;
    long long num = j - i;
    ans =
        (ans + num * ((a[i] * fac1[n - i - 1] % mod) * fac2[n - i + 1])) % mod;
  }
  cout << ans << endl;
  return 0;
}
