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 INF = 1e9 + 7;
const long long MAXN = 1e6 + 7;
const double EPS = 1e-8;
long long power(long long a, long long k) {
  if (!k) {
    return 1;
  }
  long long b = power(a, k / 2);
  b = b * b % INF;
  if (k % 2) {
    return a * b % INF;
  } else {
    return b;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  vector<long long> f(MAXN, 1);
  for (long long i = 1; i < MAXN; i++) {
    f[i] = f[i - 1] * i % INF;
  }
  long long n;
  cin >> n;
  map<long long, long long> cnt;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    x = -x;
    cnt[x]++;
  }
  long long all = 0;
  long long res = 0;
  for (auto p : cnt) {
    long long v = -p.first, c = p.second;
    all += c;
    if (all > c) {
      res = (res + c * v % INF * power(all, INF - 2)) % INF;
    }
  }
  cout << res * f[n] % INF;
}
