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 mod = 1e9 + 7;
long long inv(int x) {
  int p = mod - 2;
  long long ret = 1;
  long long now = x;
  while (p) {
    if (p & 1) {
      ret *= now;
      ret %= mod;
    }
    now *= now;
    now %= mod;
    p >>= 1;
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  long long tot = 1;
  for (int i = 1; i < n + 1; i++) {
    tot *= i;
    tot %= mod;
  }
  vector<int> v;
  for (int i = 0; i < n; i++) {
    int add;
    cin >> add;
    v.push_back(add);
  }
  sort(v.rbegin(), v.rend());
  long long cur = 0;
  long long ans = 0;
  int last = 0;
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    if (v[i] == last) {
      cur++;
      cnt++;
      continue;
    }
    if (last != v[0]) {
      ans += tot * last % mod * inv(cur) % mod * cnt % mod;
      ans %= mod;
    }
    cur++;
    cnt = 1;
    last = v[i];
  }
  if (last != v[0]) {
    ans += tot * last % mod * inv(cur) % mod * cnt % mod;
    ans %= mod;
  }
  cout << ans << '\n';
}
