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 MAX_N = 1e6 + 10;
const int mod = 1e9 + 7;
long long a[MAX_N], n, fact[MAX_N], rfact[MAX_N];
long long num(int t) { return fact[n - t - 1] * rfact[n - t + 1] % mod; }
int main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  fact[0] = rfact[n + 1] = 1;
  for (int i = 1; i <= n + 1; i++) fact[i] = (fact[i - 1] * i) % mod;
  for (int i = n; i >= 1; i--) rfact[i] = (rfact[i + 1] * i) % mod;
  sort(a + 1, a + n + 1);
  long long cnt = 0, curr = 0, ans = 0;
  for (long long i = 1; i <= n && a[i] != a[n]; i++) {
    if (a[i] == a[i - 1])
      cnt++;
    else
      curr += cnt, cnt = 1;
    ans = (ans + num(curr) * a[i] % mod) % mod;
  }
  printf("%d", ans);
}
