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 md = 1000000007;
inline int add(int a, int b) {
  a += b;
  if (a >= md) a -= md;
  return a;
}
inline int mul(int a, int b) { return (int)((long long)a * b % md); }
inline int po(int a, int b) {
  int r = 1;
  while (b) {
    if (b & 1) r = mul(a, r);
    a = mul(a, a);
    b >>= 1;
  }
  return r;
}
inline int inv(int a) { return po(a, md - 2); }
inline int di(int a, int b) { return mul(a, inv(b)); }
const int N = 1000000;
int n;
int a[N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  sort(a, a + n);
  int less = 0;
  int r = 0;
  for (int i = 0; i < n && a[i] != a[n - 1]; ++i) {
    if (i && a[i] != a[i - 1]) less = i;
    r = add(r, di(a[i], n - less));
  }
  for (int i = 2; i <= n; ++i) r = mul(r, i);
  printf("%d\n", r);
  return 0;
}
