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 mod = 1e9 + 7;
int n;
long long arr[1000001], fact[1000001];
long long power(long long a, long long p) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = ans * a % mod;
    a = a * a % mod;
    p >>= 1;
  }
  return ans;
}
long long inv(long long a) { return power(a, mod - 2); }
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%lld", arr + i);
  sort(arr, arr + n);
  fact[0] = 1;
  for (int i = 1; i <= n; i++) fact[i] = fact[i - 1] * i % mod;
  int l = 0;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    if (arr[i] == arr[n - 1]) break;
    if (i != 0 && arr[i] != arr[i - 1]) l = i;
    ans = (ans + arr[i] * fact[n] % mod * inv(n - l) % mod) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}
