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 f[1000007];
long long pw(long long a, long long b) {
  if (b == 0) return 1;
  long long r = pw(a, b / 2);
  r = (r * r) % 1000000007;
  if (b % 2) r = (r * a) % 1000000007;
  return r;
}
inline long long inv(long long k) { return pw(k, 1000000007 - 2); }
inline long long placement(long long puraton, long long notun) {
  return (f[puraton + notun] * inv(f[puraton])) % 1000000007;
}
int main(void) {
  std::ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  f[0] = 1LL;
  for (long long i = 1; i <= n; i++) f[i] = (i * f[i - 1]) % 1000000007;
  map<long long, long long> mp;
  for (long long i = 0; i < n; i++) {
    long long a;
    cin >> a;
    mp[a]++;
  }
  long long rem = n;
  long long ans = 0;
  for (pair<long long, long long> p : mp) {
    rem -= p.second;
    if (rem == 0) continue;
    long long add =
        (f[rem] * (p.second * placement(rem, p.second - 1)) % 1000000007) %
        1000000007;
    add = (add * placement(rem + p.second, n - rem - p.second)) % 1000000007;
    ans = (ans + (add * p.first) % 1000000007) % 1000000007;
  }
  cout << ans << endl;
  return 0;
}
