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;
long long fac[1000005], a[1000005], ans, mod = 1e9 + 7;
long long add(long long x) { return x >= mod ? x - mod : x; }
long long mul(long long x, long long y) { return x * y % mod; }
long long Pow(long long a, long long b) {
  long long rec = 1;
  for (; b; b >>= 1, a = mul(a, a))
    if (b & 1) rec = mul(rec, a);
  return rec;
}
int main() {
  scanf("%d", &n);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = mul(fac[i - 1], i);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  sort(a + 1, a + n + 1);
  for (int i = 1, j = 1; i <= n; i = j + 1) {
    while (a[i] == a[j + 1]) j++;
    if (j == n) break;
    ans = add(ans +
              mul(a[i], mul(j - i + 1, mul(fac[n], Pow(n - i + 1, mod - 2)))));
  }
  printf("%lld", ans);
}
