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 N = 1000005;
const int inf = 1000 * 1000 * 1000;
const int mod = 1000 * 1000 * 1000 + 7;
int n;
int a[N];
int f[N], c[N];
int binpow(int a, int b) {
  if (b == 0) return 1;
  if (b % 2) return a * 1ll * binpow(a, b - 1) % mod;
  int t = binpow(a, b / 2);
  return t * 1ll * t % mod;
}
int A(int n, int k) { return f[n] * 1ll * c[n - k] % mod; }
int main() {
  cin >> n;
  f[0] = c[0] = 1ll;
  for (int i = 1; i <= n; i++) {
    f[i] = f[i - 1] * 1ll * i % mod;
    c[i] = binpow(f[i], mod - 2);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a + 1, a + n + 1);
  int answ = 0;
  for (int i = 1; i <= n; i++) {
    int x = i, s = 0;
    while (i <= n && a[i] == a[x]) {
      s++;
      i++;
    }
    if (i == n + 1) break;
    i--;
    int sum = s * 1ll * A(n, x - 1) % mod;
    sum = sum * 1ll * f[n - x] % mod;
    sum = sum * 1ll * a[x] % mod;
    answ = (answ + sum) % mod;
  }
  cout << answ << endl;
  return 0;
}
