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;
void preprocess(void) { return; }
vector<long long> fac, a;
long long pow1(long long x, long long y) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1LL) ans = (ans * x) % 1000000007;
    x = (x * x) % 1000000007;
    y = y >> 1;
  }
  return ans;
}
long long inv(long long x) { return pow1(x, 1000000007 - 2); }
vector<int> f;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(20);
  preprocess();
  int n;
  cin >> n;
  a.resize(n);
  fac.resize(n + 1);
  fac[0] = 1;
  for (auto i = (1); i < n + 1; i++) fac[i] = (i * fac[i - 1]) % 1000000007;
  for (auto i = (0); i < n; i++) cin >> a[i];
  long long sum = 0;
  sort(a.begin(), a.end());
  ;
  f.resize(n + 1);
  for (auto i = (0); i < n; i++) {
    int pos = i;
    while (pos < n && a[pos] == a[i]) {
      f[pos] = i;
      pos++;
    }
    i = pos - 1;
  };
  for (auto i = (0); i < n; i++) {
    if (a[i] == a[n - 1]) continue;
    sum += (a[i] * inv(n - f[i])) % 1000000007;
    sum %= 1000000007;
    ;
  }
  sum = (sum * fac[n]) % 1000000007;
  cout << sum << endl;
  long long k = a[n - 1], cnt = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (a[i] == k) {
      cnt++;
    } else
      break;
  }
}
