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;
const long long N = 1e6 + 7;
long long n, f = 1, ans;
long long a[N];
map<long long, long long, greater<long long> > M;
long long fpow(long long a, long long b) {
  long long res = 1;
  for (long long i = 0; i < 30; ++i) {
    if (b & (1 << i)) {
      res *= a;
      res %= MOD;
    }
    a *= a;
    a %= MOD;
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n;
  for (long long i = 1; i <= n; ++i) {
    long long x;
    cin >> x;
    ++M[x];
    f *= i;
    f %= MOD;
  }
  long long cnt = 0;
  for (auto it : M) {
    ans += it.first * f % MOD * it.second % MOD *
           fpow(cnt + it.second, MOD - 2) % MOD;
    ans %= MOD;
    cnt += it.second;
  }
  ans -= f * (*M.begin()).first % MOD;
  ans += MOD;
  ans %= MOD;
  cout << ans << '\n';
}
