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 n, oppo[1000005], ans, a[1000005];
long long MOD = 1e9 + 7;
long long mog(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % MOD;
    b >>= 1;
    a = a * a % MOD;
  }
  return ans;
}
int main() {
  cin >> n;
  long long f = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &oppo[i]);
    a[i] = i;
    f = f * i % MOD;
  }
  sort(oppo + 1, oppo + n + 1);
  for (int i = 1; oppo[i] != oppo[n]; i++) {
    int j = i;
    if (oppo[j - 1] == oppo[j]) {
      a[i] = a[i - 1];
      j = a[i];
    }
    ans += oppo[j] * f % MOD * mog(n - (j - 1), MOD - 2) % MOD;
  }
  cout << ans % MOD << endl;
  return 0;
}
