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 N = 1e6 + 5;
const long long mod = 1e9 + 7;
long long n, a[N];
long long lfact[N], rfact[N];
long long calc(long long x) {
  return (lfact[n - x - 1] * rfact[n - x + 1]) % mod;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; ++i) cin >> a[i];
  sort(a + 1, a + n + 1);
  lfact[0] = rfact[n + 1] = 1;
  for (long long i = 1; i <= n + 1; ++i) lfact[i] = (lfact[i - 1] * i) % mod;
  for (long long i = n; i >= 0; --i) rfact[i] = (rfact[i + 1] * i) % mod;
  long long ans = 0, cur = 0, cnt = 0;
  for (long long i = 1; i <= n && a[i] != a[n]; ++i) {
    if (a[i] == a[i - 1])
      ++cnt;
    else
      cur += cnt, cnt = 1;
    ans = (ans + a[i] * calc(cur)) % mod;
  }
  cout << ans << endl;
}
