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 long long mod = 1000000007;
long long powmod(long long a, long long b) {
  if (b == 1) return a % mod;
  if (b % 2 == 1) return (a * powmod(a, b - 1)) % mod;
  long long x = powmod(a, b / 2);
  return (x * x) % mod;
}
long long _fact(long long n) {
  long long fact = 1;
  for (int i = 1; i <= n; i++) {
    fact = (fact * i) % mod;
  }
  return fact;
}
int main() {
  int n;
  cin >> n;
  int a[n];
  map<int, int> cnt;
  for (int i = 0; i < n; ++i) {
    int x;
    cin >> a[i];
  }
  sort(a, a + n);
  long long q[n];
  long long fact = _fact(n);
  long long divider;
  for (int i = 0; i < n; ++i) {
    if (i == 0 || a[i] != a[i - 1]) divider = n - i;
    q[i] = (a[i] * fact) % mod;
    q[i] = (q[i] * powmod(divider, mod - 2)) % mod;
  }
  long long ans = 0;
  for (int i = 0; i < n - 1; ++i) {
    if (a[i] == a[n - 1]) break;
    ans = (ans + q[i]) % mod;
  }
  cout << ans;
}
