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, ar[1000001], mod = 1e9 + 7, factorialn = 1;
long long ans;
long long binpow(long long a, long long b) {
  if (b == 1) return a;
  if (b % 2 == 1)
    return a * 1ll * binpow(a, b - 1) % mod;
  else {
    long long h = binpow(a, b / 2);
    return h * 1ll * h % mod;
  }
}
long long ok(long long a, long long b) {
  long long h;
  h = a * 1ll * factorialn % mod;
  return h * 1ll * binpow(b, mod - 2) % mod;
}
int main() {
  cin >> n;
  for (long long i = 1; i <= n; i++)
    cin >> ar[i], factorialn = factorialn * 1ll * i % mod;
  sort(ar + 1, ar + n + 1);
  long long x = 1;
  for (long long i = 2; i <= n + 1; i++) {
    if (ar[i] != ar[i - 1]) {
      if (n - (i - x - 1) != x)
        ans = (ans + ok(x, n - (i - x - 1)) * 1ll * ar[i - 1]) % mod;
      x = 1;
    } else
      x++;
  }
  cout << ans;
  return 0;
}
