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 double PI = 2 * acos(0.0);
void yes() { cout << "YES\n"; }
void no() {
  cout << "NO\n";
  cout << -1 << "\n";
  exit(0);
}
long long m(long long a) { return (a) % 1000000007; }
long long cel(long long x1, long long y1) {
  ;
  if ((x1 % y1) == 0) return x1 / y1;
  return x1 / y1 + 1;
}
long long power(long long a, long long b) {
  if (b == 0) return 1;
  long long d = power(a, b / 2);
  d = m(d * d);
  if (b & 1) d = m(d * a);
  return d;
}
long long modinverse(long long a) { return power(a, 1000000007 - 2); }
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  vector<long long> arr(n);
  for (long long i = 0; i < n; i++) cin >> arr[i];
  long long x = 0;
  long long fact = 1;
  for (long long i = 1; i <= n; i++) {
    fact *= i;
    fact %= 1000000007;
  }
  if (n == 1) {
    cout << 0 << "\n";
    return 0;
  }
  sort(arr.begin(), arr.end());
  long long f = n - 1;
  while (f >= 0 && arr[f] == arr[n - 1]) {
    f--;
    x++;
  }
  if (f == -1) {
    cout << 0 << "\n";
    return 0;
  }
  long long ans = 0;
  while (f >= 0) {
    long long l = f;
    while (l >= 0 && arr[l] == arr[f]) {
      l--;
      x++;
    }
    x--;
    l++;
    long long num = f - l + 1;
    long long aa = (fact * modinverse(x + 1)) % 1000000007;
    aa *= num;
    aa %= 1000000007;
    aa *= arr[l];
    ans += aa;
    ans %= 1000000007;
    f = l - 1;
    x++;
  }
  cout << ans << "\n";
}
