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 int N = 1E6 + 5, M = 1E9 + 7;
int a[N], n;
long long fact[N];
long long fpow(int b, int e) {
  if (e == 0) return 1;
  if (e == 1) return b;
  long long r = fpow(b, e / 2);
  r = r * r % M;
  if (e & 1)
    return r * b % M;
  else
    return r;
}
int main() {
  int i, j;
  long long ans = 0;
  scanf("%d", &n);
  fact[0] = 1;
  for (i = 1; i <= n; i++) fact[i] = fact[i - 1] * i % M;
  for (i = 0; i < n; i++) scanf("%d", a + i);
  sort(a, a + n);
  for (i = 0; i < n; i = j) {
    for (j = i + 1; j < n && a[j] == a[i]; j++)
      ;
    if (j == n) break;
    ans = (ans + a[i] * (j - i) % M * fact[n] % M * fpow(n - i, M - 2)) % M;
  }
  printf("%I64d", ans);
  return 0;
}
