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, a[1000001], p = 1e9 + 7, ans;
long long ksm(long long x, long long y) {
  long long xlh = 1;
  while (y) {
    if (y & 1) xlh = xlh * x % p;
    x = x * x % p;
    y /= 2;
  }
  return xlh;
}
int main() {
  long long i, j, now;
  scanf("%lld", &n);
  for (i = 1; i <= n; i++) scanf("%lld", &a[i]);
  sort(a + 1, a + n + 1);
  now = 1;
  for (i = 1; i <= n; i++) now = now * i % p;
  for (i = 1; i <= n; i = j) {
    j = i;
    while (a[i] == a[j] && j <= n) j++;
    if (j <= n)
      ans =
          (ans + now * ksm(n - i + 1, p - 2) % p * (j - i) % p * a[i] % p) % p;
  }
  printf("%lld", ans);
}
