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 = 1010000;
const int MOD = 1e9 + 7;
int n, m, a[MAXN];
long long ksm(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  long long t = ksm(x, y / 2);
  t = t * t % MOD;
  if (y & 1) t = t * x % MOD;
  return t;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), m = max(m, a[i]);
  sort(a + 1, a + 1 + n);
  long long ans = 0, last = 0, fac = 1;
  for (int i = 2; i <= n; i++) fac = fac * i % MOD;
  for (int i = 1; i <= n; i++) {
    if (m == a[i]) break;
    if (a[i] != a[i - 1]) last = fac * ksm(n - i + 1, MOD - 2) % MOD;
    ans = (ans + last * a[i]) % MOD;
  }
  printf("%lld", ans);
  return 0;
}
