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;
int n;
int a[1000005];
vector<int> v;
map<int, int> mp;
map<int, bool> used;
long long power(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  if (y % 2 == 0) return power(x * x % 1000000007, y / 2);
  return power(x * x % 1000000007, y / 2) * x % 1000000007;
}
long long inverse(long long k) { return power(k, 1000000007 - 2); }
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    mp[a[i]]++;
  }
  sort(a, a + n);
  long long ans = 0;
  for (int i = 0; i < n - 1; i++) {
    if (!used[a[i]] && a[i] != a[n - 1]) {
      used[a[i]] = 1;
      ans = (ans + mp[a[i]] * a[i] * inverse(n - i)) % 1000000007;
    }
  }
  long long stair = 1;
  for (int i = 1; i <= n; i++) {
    stair = (stair * i) % 1000000007;
  }
  ans = (ans * stair) % 1000000007;
  printf("%lld", ans);
  return 0;
}
