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 infl = 1e18 + 5;
long long n, m, k, q, x, y, val, t, i, j;
long long ind, cnt, sz, sm, ans, mx, mn;
long long a[1000004], f[1000004];
long long modpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res *= a;
      res %= 1000000007;
    }
    b /= 2;
    a *= a;
    a %= 1000000007;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  if (fopen("inp.txt", "r")) {
    freopen("myfile.txt", "w", stdout);
    freopen("inp.txt", "r", stdin);
  }
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  f[0] = 1;
  for (i = 1; i < n + 1; i++) f[i] = (i * f[i - 1]) % 1000000007;
  sort(a, a + n);
  long long l = 0, eq = 0, prev = -infl;
  for (i = 0; i < n; i++) {
    if (prev < a[i]) {
      prev = a[i];
      l = l + eq;
      eq = 1;
    } else
      eq++;
    if (a[i] == a[n - 1]) break;
    ans += (f[n] * ((a[i] * modpow(n - l, 1000000007 - 2)) % 1000000007)) %
           1000000007;
    ans %= 1000000007;
  }
  cout << ans;
  return 0;
}
