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 long long yu = 1000000007;
long long n, jie[1000100], ni[1000100], a[1000100];
long long ksm(long long a, long long b) {
  long long an = 1;
  while (b) {
    if (b & 1) {
      an = (an * a) % yu;
    }
    a = (a * a) % yu;
    b >>= 1;
  }
  return an;
}
long long cc(int a, int b) {
  if (b == 0 || a == b) return 1;
  return ((jie[a] * ni[b]) % yu * ni[a - b]) % yu;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  long long ans = 0;
  jie[0] = 1;
  jie[1] = 1;
  for (int i = 2; i <= 1000000; i++) {
    jie[i] = (jie[i - 1] * i) % yu;
  }
  ni[1000000] = ksm(jie[1000000], yu - 2);
  for (int i = 999999; i >= 1; i--) {
    ni[i] = (ni[i + 1] * (i + 1)) % yu;
  }
  sort(a + 1, a + 1 + n);
  int mx = a[n];
  long long q = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == mx) break;
    if (a[i] != a[i - 1]) q = jie[i - 1] * jie[n - i] % yu * cc(n, i - 1) % yu;
    ans = (ans + a[i] * q % yu) % yu;
  }
  printf("%lld\n", ans);
  return 0;
}
