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>
const int inf = (1ll << 30) - 1;
const int maxn = (int)1e5 + 10;
using namespace std;
long long fact[2000100];
int mod = 1000 * 1000 * 1000 + 7;
int n;
int a[2000100];
long long binpow(long long x, int y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return res;
}
long long C(int n, int k) {
  return fact[n] * binpow(fact[n - k] * 1ll * fact[k] % mod, mod - 2) % mod;
}
void solve() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a, a + n);
  fact[0] = 1;
  for (long long i = 1; i <= 2 * n; i++) {
    fact[i] = fact[i - 1] * i % mod;
  }
  long long ans = 0;
  for (int i = 0; i < n;) {
    int j = i;
    while (i < n && a[i] == a[j]) i++;
    if (i != n) {
      ans = (ans + ((((i - j) * 1ll * C(n, j)) % mod * fact[j] % mod) * 1ll *
                    a[j] % mod) *
                       fact[n - j - 1] % mod) %
            mod;
    }
  }
  cout << ans << endl;
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}
