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 = 1000010, M = 1e9 + 7;
int n, ans, fac[N], inv[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = 1ll * fac[i - 1] * i % M;
  inv[0] = inv[1] = 1;
  for (int i = 2; i <= n; ++i) inv[i] = 1ll * (M - M / i) * inv[M % i] % M;
  map<int, int> s;
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    ++s[x];
  }
  int k(0);
  for (map<int, int>::reverse_iterator i = s.rbegin(); i != s.rend(); ++i) {
    k += i->second;
    if (i != s.rbegin())
      (ans += 1ll * fac[n] * inv[k] % M * (i->second) % M * (i->first) % M) %=
          M;
  }
  cout << ans;
}
