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 MOD = 1e9 + 7;
const int MAXN = 1000009;
int n;
long long a[MAXN], inv[MAXN], l[MAXN];
void get_inv() {
  inv[1] = 1;
  for (int i = 2; i < MAXN; ++i) {
    inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;
  }
}
int main() {
  scanf("%d", &n);
  get_inv();
  for (int i = 0; i < n; ++i) scanf("%lld", &a[i]);
  sort(a, a + n);
  long long x = 1, ans = 0;
  for (long long i = 2; i <= n; ++i) {
    x = (x * i) % MOD;
  }
  l[0] = 0;
  for (int i = 0; i < n && a[i] != a[n - 1]; ++i) {
    if (i != 0) {
      if (a[i] > a[i - 1])
        l[i] = i;
      else
        l[i] = l[i - 1];
    }
    ans = (ans + (a[i] * x % MOD) * inv[n - l[i]] % MOD) % MOD;
  }
  printf("%lld\n", ans);
  return 0;
}
