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 MAX_N = 1000005;
long long inv[MAX_N], fac[MAX_N], finv[MAX_N];
void make() {
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (int i = 2; i < MAX_N; i++) {
    inv[i] = 1000000007 - inv[1000000007 % i] * (1000000007 / i) % 1000000007;
    fac[i] = fac[i - 1] * (long long)i % 1000000007;
    finv[i] = finv[i - 1] * inv[i] % 1000000007;
  }
}
long long add(long long x, long long y) { return (x + y) % 1000000007; }
long long sub(long long x, long long y) {
  return (x + 1000000007 - y) % 1000000007;
}
long long mul(long long x, long long y) { return x * y % 1000000007; }
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> vec(n);
  for (int i = 0; i < (int)(n); ++i) {
    cin >> vec[i];
  }
  sort((vec).begin(), (vec).end());
  make();
  int cnt = 0;
  int kp = 0;
  long long ans = 0;
  int mx = vec[n - 1];
  for (int i = 0; i < (int)(n); ++i) {
    if (vec[i] == mx) {
      break;
    }
    if (i == 0 || vec[i - 1] < vec[i]) {
      cnt += kp;
      kp = 1;
    } else {
      kp++;
    }
    ans = add(ans, mul(mul(fac[n], inv[n - cnt]), vec[i]));
  }
  cout << ans << "\n";
  return 0;
}
