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;
long long mod = 1e9 + 7;
long long modexp(long long x, long long n) {
  if (n == 0) return 1;
  if (n % 2 == 0) {
    long long y = modexp(x, n / 2) % mod;
    return (y * y) % mod;
  }
  return (x * modexp(x, n - 1) % mod) % mod;
}
long long powr(long long x, long long n) {
  long long ans = 1;
  for (int i = 1; i <= n; i++) ans = ans * x;
  return ans;
}
long long fact[1000005];
map<long long, long long> m;
int main() {
  fact[0] = 1;
  for (long long i = 1; i <= 1000000; i++) {
    fact[i] = (fact[i - 1] * i) % mod;
  }
  long long n;
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) {
    long long x;
    scanf("%lld", &x);
    m[x]++;
  }
  long long curr = 0, ans = 0;
  int k = m.size(), cnt = 0;
  for (map<long long, long long>::iterator it = m.begin(); it != m.end();
       it++) {
    cnt++;
    long long x = it->first;
    long long u = it->second;
    curr += u;
    long long v = n - curr;
    long long contri = (fact[n] * x) % mod;
    contri = (contri * u) % mod;
    contri = (contri * (modexp(u + v, mod - 2))) % mod;
    if (cnt != k) ans = (ans + contri) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}
