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 int INF = (1 << 30) - 1;
const long long LINF = (1LL << 60) - 1;
template <class T>
bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  } else
    return false;
}
template <class T>
bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  } else
    return false;
}
long long mod_pow(long long a, long long n, long long p) {
  long long ans = 1, tmp = a % p;
  while (n > 0) {
    if (n & 1) ans = ans * tmp % p;
    tmp = tmp * tmp % p;
    n = (n >> 1);
  }
  return ans;
}
int main() {
  int n;
  cin >> n;
  vector<long long> a(n), kai(n + 1, 1);
  for (int i = 0; i < n; i++) cin >> a[i];
  long long mod = 1e9 + 7, ans = 0;
  for (long long x = 1; x <= n; x++) kai[x] = kai[x - 1] * x % mod;
  sort(a.begin(), a.end());
  for (int i = 0; i < n; i++) {
    int j = upper_bound(a.begin(), a.end(), a[i]) - a.begin();
    if (a[i] == a[n - 1]) break;
    long long S = i, L = n - 1 - i;
    ans = (ans + a[i] * kai[n] % mod * mod_pow(L + 1, mod - 2, mod) % mod *
                     (j - i) % mod) %
          mod;
    i = j - 1;
  }
  cout << ans << endl;
  return 0;
}
