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 MOD = 1e9 + 7, N = 1e6 + 11;
long long f[N], fi[N];
void pre() {
  f[0] = f[1] = fi[0] = fi[1] = 1;
  for (long long i = 2; i < N; i++) f[i] = i * f[i - 1] % MOD;
  for (long long i = 2; i < N; i++) fi[i] = (MOD - MOD / i) * fi[MOD % i] % MOD;
  for (long long i = 2; i < N; i++) fi[i] = fi[i] * fi[i - 1] % MOD;
}
long long c(long long n, long long k) {
  return f[n] * fi[k] % MOD * fi[n - k] % MOD;
}
long long a[1000006];
int32_t main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  ;
  long long n, ans = 0;
  cin >> n;
  pre();
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  for (long long i = 0, j = 0; i < n; i = j) {
    long long s = 0;
    for (; a[j] == a[i]; j++) s += a[j];
    if (j < n) ans += s % MOD * f[i] % MOD * f[n - 1 - i] % MOD * c(n, i) % MOD;
  }
  cout << ans % MOD << endl;
}
