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 p = 1e9 + 7;
const int nn = 1e6 + 5;
long long n, a[nn], jc[nn], jm[nn], ans;
long long re() {
  char c = getchar();
  long long x = 0;
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) x = x * 10 + (c ^ 48), c = getchar();
  return x;
}
long long qs(long long x, long long y) {
  long long z = 1;
  while (y) {
    if (y & 1) z = z * x % p;
    x = x * x % p, y >>= 1;
  }
  return z;
}
int main() {
  n = re();
  jc[0] = 1;
  for (long long i = 1; i <= n; ++i) jc[i] = jc[i - 1] * i % p;
  jm[n] = qs(jc[n], p - 2);
  for (long long i = n - 1; i >= 0; --i) jm[i] = jm[i + 1] * (i + 1) % p;
  for (long long i = 1; i <= n; ++i) a[i] = re();
  sort(a + 1, a + n + 1);
  long long sum = 0;
  for (long long i = 1; i < n; ++i) {
    ++sum;
    if (a[i] == a[i + 1]) continue;
    ans = (ans + a[i] * sum % p * jc[n] % p * jc[n - i + sum - 1] % p *
                     jm[n - i + sum]) %
          p,
    sum = 0;
  }
  printf("%lld", ans);
  return 0;
}
