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 n;
int a[1000005];
long long ans = 0;
void exgcd(int a, int b, int &x, int &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return;
  }
  exgcd(b, a % b, y, x);
  y -= (a / b) * x;
}
int inv(int a) {
  int x, y;
  exgcd(a, 1000000007, x, y);
  return (x % 1000000007 + 1000000007) % 1000000007;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  long long f = 1;
  for (int i = 1; i <= n; i++) f = f * i % 1000000007;
  sort(a + 1, a + 1 + n);
  int nex = 1;
  for (int i = 1; i <= n; i = nex) {
    if (a[i] == a[n]) break;
    while (nex <= n && a[i] == a[nex]) nex++;
    ans = (ans + f * inv(n - i + 1) % 1000000007 * a[i] % 1000000007 *
                     (nex - i) % 1000000007) %
          1000000007;
  }
  cout << ans << endl;
}
