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;
int n, x, p = 1000000007;
map<int, int> g;
int v[1000020];
int f[1000020];
long long z = 0;
int main() {
  scanf("%d", &n);
  v[0] = v[1] = 1;
  for (int i = 2; i <= n; i++) {
    v[i] = (long long)v[p % i] * (p - p / i) % p;
  }
  f[0] = 1;
  for (int i = 1; i <= n; i++) {
    f[i] = (long long)f[i - 1] * i % p;
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    g[x]++;
  }
  long long z = 0, c = n;
  for (pair<int, int> i : g) {
    if (c == i.second) {
      break;
    }
    z = (z + (long long)f[n] * i.second % p * i.first % p * v[c] % p) % p;
    c -= i.second;
  }
  printf("%lld\n", z);
  return 0;
}
