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 po(long long int base) {
  long long int power = 1e9 + 5;
  long long int mod = 1e9 + 7;
  long long int ans = 1;
  while (power > 0) {
    if (power % 2 == 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    power /= 2;
  }
  return ans;
}
long long int fact[1000005] = {0}, mod = 1e9 + 7, a[1000005] = {0},
              freq[1000005] = {0};
int main() {
  ios::sync_with_stdio(false);
  fact[0] = 1;
  for (long long int i = 1; i <= 1000000; i++)
    fact[i] = (fact[i - 1] * i) % mod;
  int n;
  cin >> n;
  int m = n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  int cnt = 1;
  vector<pair<long long int, long long int> > v;
  for (int i = 1; i < n; i++) {
    if (a[i] == a[i - 1])
      cnt++;
    else {
      v.push_back({a[i - 1], cnt});
      cnt = 1;
    }
  }
  v.push_back({a[n - 1], cnt});
  freq[0] = v[0].second;
  n = v.size();
  for (int i = 1; i < n; i++) freq[i] = freq[i - 1] + v[i].second;
  int curr = v[n - 1].second;
  long long int ans = 0;
  for (int i = n - 2; i >= 0; i--) {
    long long int d = curr + v[i].second;
    long long int temp = (fact[m] * po(fact[d])) % mod;
    temp = (temp * fact[d - 1]) % mod;
    temp = (temp * v[i].second) % mod;
    ans = (ans + (temp * v[i].first) % mod) % mod;
    curr += v[i].second;
  }
  cout << ans;
  return 0;
}
