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;
int Mod = 1000000007, F[1010000], InvF[1010000], w[1010000];
int Pow(int a, int b) {
  int r = 1;
  while (b) {
    if (b & 1) r = r * 1ll * a % Mod;
    a = 1ll * a * a % Mod;
    b >>= 1;
  }
  return r;
}
int n;
int main() {
  int i;
  F[0] = InvF[0] = 1;
  for (i = 1; i <= 1000000; i++) {
    F[i] = 1ll * F[i - 1] * i % Mod;
  }
  InvF[i - 1] = Pow(F[i - 1], Mod - 2);
  for (i = i - 1; i >= 1; i--) {
    InvF[i - 1] = 1ll * InvF[i] * i % Mod;
  }
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &w[i]);
  sort(w + 1, w + n + 1);
  int pv = 0;
  long long res = 0;
  for (i = 1; i <= n; i++) {
    while (w[pv + 1] < w[i]) pv++;
    res = (res + 1ll * F[n] * InvF[n - pv] % Mod * F[n - pv - 1] % Mod * w[i]) %
          Mod;
  }
  res = (res - 1ll * F[n] * w[n] % Mod + Mod) % Mod;
  printf("%lld\n", res);
}
