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 = 1000000 + 5;
const int p = 1e9 + 7;
long long f[maxn], a[maxn];
long long mog(long long a, int b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % p;
    b >>= 1;
    a = a * a % p;
  }
  return ans;
}
int n;
int main() {
  cin >> n;
  f[0] = 1;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    f[i] = f[i - 1] * i % p;
  }
  sort(a + 1, a + 1 + n);
  int END = a[n];
  long long ans = 0;
  int x = 0;
  for (int i = 1; i <= n && a[i] != END; i++) {
    if (a[i] != a[i - 1]) x = i - 1;
    ans = (ans + a[i] * f[n] % p * mog(n - x, p - 2)) % p;
  }
  cout << ans % p;
  return 0;
}
