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 MOD = 1E9 + 7;
int fac[1000010], n, a[1000010], ans;
int mypow(int x, int n) {
  int ans = 1;
  for (; n; n >>= 1, x = 1ll * x * x % MOD)
    if (n & 1) ans = 1ll * ans * x % MOD;
  return ans;
}
int cal(int x) { return 1ll * fac[n] * mypow(x + 1, MOD - 2) % MOD; }
int main() {
  fac[0] = 1;
  for (int i = 1; i <= 1000000; i++) {
    fac[i] = 1ll * fac[i - 1] * i % MOD;
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n;) {
    if (a[i] == a[n]) {
      break;
    }
    int j = i;
    for (; j <= n && a[j] == a[i];) {
      j++;
    }
    ans = (ans + 1ll * cal(n - i) * a[i] % MOD * (j - i)) % MOD;
    i = j;
  }
  printf("%d\n", ans);
  return 0;
}
