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 = (int)1e9 + 7;
inline int mul(int a, int b) { return ((long long)a * b) % mod; }
inline int add(int a, int b) { return (a + b) % mod; }
int bin_pow(int a, int n) {
  int res = 1;
  while (n) {
    if (n & 1) res = mul(res, a);
    a = mul(a, a);
    n >>= 1;
  }
  return res;
}
map<int, int> m;
int main() {
  int n;
  scanf("%d", &n);
  int fac = 1;
  for (int i = 2; i <= n; ++i) fac = mul(fac, i);
  int x;
  int mx = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &x);
    m[x]++;
    mx = max(mx, x);
  }
  int ans = 0;
  int cur = 0;
  for (auto &j : m) {
    if (j.first == mx) break;
    ans = add(ans, mul(j.first, mul(j.second, bin_pow(n - cur, mod - 2))));
    cur += j.second;
  }
  cout << mul(ans, fac) << endl;
  return 0;
}
