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 MAX = 2e6 + 7, P = 1e9 + 7;
int f[MAX], a[MAX], n;
int fpow(int b, int p) {
  long long int r = 1ll;
  while (p) {
    if (p & 1) r *= b, r %= P;
    b = 1ll * b * b % P;
    p >>= 1;
  }
  return int(r);
}
void pre() {
  f[0] = 1;
  for (int i = (1); i <= (n); ++i) {
    cin >> a[i];
    f[i] = 1ll * f[i - 1] * i % P;
  }
  sort(a + 1, a + n + 1);
}
void solve() {
  int ans = 0, lst = 0;
  for (int i = (1); i <= (n); ++i) {
    if (a[i] == a[n]) continue;
    if (i > 1 && a[i] != a[i - 1]) lst = i - 1;
    int cur = 1ll * a[i] * f[n] % P;
    cur = 1ll * cur * fpow(n - lst, P - 2) % P;
    ans += cur;
    if (ans >= P) ans -= P;
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n;
  pre();
  solve();
  return 0;
}
