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 expo(long long base, long long exponent, long long mod) {
  long long ans = 1;
  while (exponent != 0) {
    if ((exponent & 1) == 1) {
      ans = ans * base;
      ans = ans % mod;
    }
    base = base * base;
    base %= mod;
    exponent >>= 1;
  }
  return ans % mod;
}
long long n, a[1000006], f[1000006];
void solve() {
  long long ans = 0;
  f[0] = 1;
  for (long long i = 1; i < 1000003; i++) {
    f[i] = (i * f[i - 1]) % 1000000007;
  }
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  long long i = 1;
  while (i < n) {
    long long cnt = 1;
    while (i < n && a[i] == a[i - 1]) i++, cnt++;
    if (i - 1 == n - 1) break;
    long long cur = (a[i - 1] * cnt) % 1000000007;
    cur = (cur * expo(n - i + cnt, 1000000007 - 2, 1000000007)) % 1000000007;
    ans = (ans + cur) % 1000000007;
    i++;
  }
  cout << (f[n] * ans) % 1000000007;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}
