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, a[1000005], f = 1, inv[1000005], tot[1000005], tut[1000005], all, ans;
int fe(int x, int y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  int res = fe(x, y / 2);
  res = 1ll * res * res % 1000000007;
  if (y & 1) res = 1ll * res * x % 1000000007;
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i <= n; i++) {
    inv[i] = fe(i, 1000000007 - 2);
    f = 1ll * f * i % 1000000007;
  }
  sort(a + 1, a + 1 + n);
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    ++cnt;
    while (i + 1 <= n && a[i] == a[i + 1]) {
      tot[cnt]++;
      i++;
    }
    tut[cnt] = a[i];
    tot[cnt]++;
  }
  all = n;
  for (int i = 1; i < cnt; i++) {
    ans = (ans + 1ll * tut[i] * tot[i] * inv[all] % 1000000007) % 1000000007;
    all -= tot[i];
  }
  printf("%d", 1ll * ans * f % 1000000007);
}
