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;
long long ModExp(int a, int n) {
  long long y = 1, x = a;
  while (n) {
    if (n % 2) y = (y * x) % 1000000007;
    x = (x * x) % 1000000007;
    n /= 2;
  }
  return y;
}
long long ModInv(int a) { return ModExp(a, 1000000007 - 2); }
int main() {
  int n;
  vector<int> a;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int j;
    scanf("%d", &j);
    a.push_back(j);
  }
  sort(a.begin(), a.end());
  long long ans = 0;
  for (int i = 0, p = 0; i < n;) {
    int u = 1;
    for (++i; i < n && a[i] == a[i - 1]; ++i, ++u)
      ;
    p += u;
    int v = n - p;
    if (v != 0) {
      long long ansx = (u * ModInv(u + v)) % 1000000007;
      ans = (ans + a[i - 1] * ansx) % 1000000007;
    }
  }
  for (int i = 1; i <= n; ++i) ans = (ans * i) % 1000000007;
  printf("%lld\n", ans);
  return 0;
}
