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 int N = 100005;
const int mod = 1e9 + 7;
const long long inf = 1e18;
int bin_pow(int a, int n) {
  int res = 1;
  while (n) {
    if (n & 1) res = (res * 1ll * a) % mod;
    n >>= 1;
    a = (a * 1ll * a) % mod;
  }
  return res;
}
int inv(int a) { return bin_pow(a, mod - 2); }
map<int, int> a;
int n;
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    a[x]++;
  }
  int ans = 0;
  int pr_cnt = 0;
  for (auto it : a) {
    int cnt = it.second;
    if (pr_cnt + cnt != n)
      ans += ((cnt * 1ll * inv(n - pr_cnt) % mod) * 1ll * it.first) % mod;
    ans %= mod;
    pr_cnt += cnt;
  }
  for (int i = 1; i <= n; ++i) {
    ans = (ans * 1ll * i) % mod;
  }
  cout << ans << endl;
  return 0;
}
