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 = 1000000007;
long long pw(long long b, long long n, long long m = mod, long long a = 1) {
  while (n) {
    if (n & 1) a = a * b % m;
    b = b * b % m;
    n >>= 1;
  }
  return a;
}
long long a[1000006], fac[1000006], ifac[1000006], inv[1000006];
map<long long, int> cnt;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = fac[i - 1] * i % mod;
  ifac[n] = pw(fac[n], mod - 2);
  for (int i = n - 1; i >= 0; --i) ifac[i] = ifac[i + 1] * (i + 1) % mod;
  for (int i = 2; i <= n; ++i) inv[i] = ifac[i] * fac[i - 1] % mod;
  for (int i = 0; i < n; ++i) cin >> a[i], ++cnt[a[i]];
  sort(a, a + n);
  int nn = unique(a, a + n) - a;
  long long ans = 0;
  int all = n;
  for (int i = 0; i < nn - 1; ++i) {
    int app = cnt[a[i]];
    ans += app * inv[all] % mod * fac[n] % mod * a[i] % mod;
    all -= app;
  }
  cout << ans % mod << endl;
}
