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 mod = 1e9 + 7;
int a[1000005];
long long f[1000005];
long long rf[1000005];
map<int, int> ht;
map<int, int>::iterator its;
long long mul(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b % 2 == 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b /= 2;
  }
  return res;
}
long long ncr(long long a, long long b) {
  return f[a] * rf[b] % mod * rf[a - b] % mod;
}
int main() {
  f[0] = 1;
  rf[0] = 1;
  for (int i = 1; i <= 1000000; i++)
    f[i] = (f[i - 1] * i) % mod, rf[i] = mul(f[i], mod - 2);
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), ht[a[i]]++;
  int sum = 0;
  long long ans = 0;
  for (its = ht.begin(); its != ht.end(); its++) {
    int l = sum, r = n - sum - its->second;
    if (r != 0)
      ans = (ans + ncr(its->second, 1) * f[r + its->second - 1] % mod *
                       ncr(n, l) % mod * f[l] % mod * its->first % mod) %
            mod;
    sum += its->second;
  }
  printf("%lld\n", ans);
}
