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 MAXN = 1000010;
const long long MOD = 1000000000 + 7;
int N;
long long ar[MAXN];
long long factorial[MAXN];
long long lesser[MAXN];
void input() {
  cin >> N;
  for (int i = 0; i < N; i++) cin >> ar[i];
}
long long raisePower(long long base, long long exp) {
  if (exp == 0) return 1;
  if (exp == 1) return base;
  if (exp % 2 == 1) {
    return ((base * raisePower(base, exp - 1)) % MOD);
  } else {
    long long tmp = raisePower(base, exp / 2);
    return ((tmp * tmp) % MOD);
  }
}
void process() {}
void solve() {
  sort(ar, ar + N);
  long long ans = 0;
  int p = 0;
  while (p < N) {
    int q = p + 1;
    while (q < N && ar[p] == ar[q]) q++;
    if (q == N) break;
    ans += ((((q - p) * ar[p]) % MOD) * raisePower(N - p, MOD - 2)) % MOD;
    ans %= MOD;
    p = q;
  }
  for (int i = 2; i <= N; i++) ans = (ans * i) % MOD;
  cout << ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  input();
  process();
  solve();
  return 0;
}
