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 + 6;
const long long mod = 1e9 + 7;
int n;
long long a[N];
long long up[N], inv[N], down[N];
void init() {
  inv[1] = down[0] = up[0] = 1;
  for (int i = 2; i <= 1e6; i++) {
    inv[i] = (mod - mod / i) * inv[mod % i] % mod;
  }
  for (int i = 1; i <= 1e6; i++) {
    up[i] = up[i - 1] * i % mod;
    down[i] = down[i - 1] * inv[i] % mod;
  }
}
int main() {
  init();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + 1 + n);
  long long ans = 0;
  for (int i = 1, j; i <= n; i = j + 1) {
    j = i;
    while (a[j + 1] == a[i]) j++;
    if (j < n)
      ans = (ans + a[i] * (j - i + 1) % mod * up[n] % mod * up[n - i] % mod *
                       up[i - 1] % mod * down[n - i + 1] % mod * down[i - 1] %
                       mod) %
            mod;
  }
  cout << ans << endl;
}
