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 oo = 1e9 + 7;
const int N = 1000010;
long long inv[N], fac_n, ans;
int n, same, a[N];
int main() {
  scanf("%d", &n);
  fac_n = inv[1] = 1;
  for (int i = 2; i <= n; ++i) {
    inv[i] = (long long)(oo - oo / i) * inv[oo % i] % oo;
    fac_n = fac_n * i % oo;
  }
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; ++i) {
    if (a[i] == a[n]) break;
    if (a[i] == a[i - 1])
      ++same;
    else
      same = 0;
    ans += fac_n * a[i] % oo * inv[n - (i - 1 - same)] % oo;
  }
  printf("%lld\n", (ans % oo + oo) % oo);
  return 0;
}
