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 a[1000010], inv[1000010], d[1000010], q[1000010];
const int mod = 1e9 + 7;
int main() {
  int n, i;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  int f = 1;
  for (i = 1; i <= n; i++) f = f * 1ll * i % mod;
  for (i = 1; i <= n; i++)
    if (a[i] == a[i - 1])
      d[i] = d[i - 1];
    else
      d[i] = i;
  inv[1] = 1;
  for (i = 2; i <= n; i++)
    inv[i] = (mod / i * 1ll * (mod - inv[mod % i])) % mod;
  int ans = 0;
  for (i = 1; i <= n; i++) q[i] = f * 1ll * inv[n - i + 1] % mod;
  for (i = 1; i <= n; i++)
    if (a[i] < a[n]) ans = (ans + q[d[i]] * 1ll * a[i]) % mod;
  printf("%d\n", ans);
  return 0;
}
