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;
long long int n, a[1000010], inv[1000010], s[1000010];
long long int pw(long long int k, long long int x) {
  if (x == 0) return 1;
  if (x == 1) return k;
  if (x % 2) {
    long long int tmp = pw(k, x / 2);
    return tmp * ((tmp * k) % 1000000007) % 1000000007;
  } else {
    long long int tmp = pw(k, x / 2);
    return (tmp * tmp) % 1000000007;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  long long int tmp = 1;
  for (long long int i = 1; i <= n; i++) tmp = (tmp * i) % 1000000007;
  for (int i = 1; i < n; i++) {
    if (a[i] != a[i - 1])
      s[i] = i;
    else
      s[i] = s[i - 1];
  }
  long long int ans = 0;
  for (long long int i = 1; i <= n; i++) inv[i] = pw(i, 1000000007 - 2);
  for (int i = 0; i < n; i++) {
    if (a[i] == a[n - 1]) break;
    ans = (ans + (((tmp * a[i]) % 1000000007) * inv[n - s[i]])) % 1000000007;
  }
  cout << ans << '\n';
  return 0;
}
