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;
constexpr int MOD = 1e9 + 7;
int n;
int board[1000000];
int l[1000000];
long long fact[1000000];
long long rfact[1000001];
int main(void) {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &board[i]);
  sort(board, board + n);
  l[0] = 0;
  for (int i = 1; i < n; i++) {
    if (board[i] == board[i - 1])
      l[i] = l[i - 1];
    else
      l[i] = i;
  }
  fact[0] = 1;
  for (int i = 1; i < n; i++) fact[i] = (fact[i - 1] * (i + 1)) % MOD;
  rfact[n - 1] = n;
  for (int i = n - 2; i >= 0; i--) rfact[i] = (rfact[i + 1] * (i + 1)) % MOD;
  rfact[n] = 1;
  long long result = 0;
  for (int i = 0; board[i] != board[n - 1]; i++) {
    result = (result +
              ((board[i] * fact[n - l[i] - 2]) % MOD * rfact[n - l[i]]) % MOD) %
             MOD;
  }
  printf("%lld\n", result);
  return 0;
}
