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>
const long long mod = 1e9 + 7;
using namespace std;
inline long long getint() {
  char ch = getchar();
  long long res = 0, sign = 1;
  if (ch == '-') sign = -1;
  while (ch < '0' || '9' < ch) {
    ch = getchar();
    if (ch == '-') sign = -1;
  }
  while ('0' <= ch && ch <= '9') {
    res = res * 10 + ch - '0';
    ch = getchar();
  }
  return res * sign;
}
long long exp_mod(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
long long n, a[1000007];
int main() {
  n = getint();
  long long top = 1;
  for (long long i = 1; i <= n; i++) top = top * i % mod;
  for (long long i = 1; i <= n; i++) a[i] = getint();
  sort(a + 1, a + 1 + n);
  long long ans = 0;
  int np;
  for (long long i = 1; i <= n; i = np) {
    np = i;
    while (a[np] == a[i] && np <= n) np++;
    if (np > n) break;
    long long dlt =
        top * exp_mod(n - i + 1, mod - 2) % mod * (np - i) % mod * a[i] % mod;
    ans = (ans + dlt) % mod;
  }
  printf("%lld\n", ans);
}
