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 INF = 3e18 + 9237;
const long long MAXN = 1e6 + 7;
const long long mod = 1e9 + 7;
const long long sqN = 320;
const long long lgN = 20;
long long a[MAXN], fct[MAXN], inv[MAXN];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  fct[0] = inv[n + 1] = 1;
  for (long long i = 1; i < n + 2; i++) {
    fct[i] = (fct[i - 1] * i) % mod;
  }
  for (long long i = n; i > 0; i--) {
    inv[i] = (inv[i + 1] * i) % mod;
  }
  sort(a + 1, a + n + 1);
  long long t = 0, cur = 0, ans = 0;
  for (long long i = 1; i <= n && a[i] != a[n]; i++) {
    if (a[i] == a[i - 1])
      cur++;
    else
      t += cur, cur = 1;
    long long temp = (fct[n - t - 1] * inv[n - t + 1]) % mod;
    ans = (ans + (temp * a[i]) % mod) % mod;
  }
  cout << ans;
}
