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 maxn = (int)1e6 + 1, mod = (int)1e9 + 7;
int n, a[maxn], inv[maxn], ans;
int main() {
  scanf("%d%d", &n, a + 1);
  inv[1] = 1;
  for (int i = 2; i <= n; ++i) {
    scanf("%d", a + i);
    inv[i] = mod - (int)(mod / i * (long long)inv[mod % i] % mod);
  }
  sort(a + 1, a + n + 1);
  for (int i = n, j = i; i >= 1; i = j) {
    for (; j >= 1 && a[i] == a[j]; --j)
      ;
    if (i < n) ans = (ans + (long long)a[i] * (i - j) % mod * inv[n - j]) % mod;
  }
  for (int i = 2; i <= n; ++i) ans = (long long)ans * i % mod;
  printf("%d\n", ans);
  return 0;
}
