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;
inline int read() {
  int sum = 0, p = 1;
  char ch = getchar();
  while (!(('0' <= ch && ch <= '9') || ch == '-')) ch = getchar();
  if (ch == '-') p = -1, ch = getchar();
  while ('0' <= ch && ch <= '9') sum = sum * 10 + ch - 48, ch = getchar();
  return sum * p;
}
const long long mod = 1e9 + 7;
const long long maxn = 1e6 + 20;
long long n, a[maxn], inv[maxn], jc[maxn], jcn[maxn], pre[maxn];
void init() {
  n = read();
  for (long long i = (1), _end_ = (n); i <= _end_; i++) a[i] = read();
  sort(a + 1, a + n + 1);
  jc[0] = jc[1] = jcn[0] = jcn[1] = inv[1] = 1;
  for (long long i = (2), _end_ = (n); i <= _end_; i++)
    inv[i] = (long long)(mod - mod / i) * inv[mod % i] % mod,
    jc[i] = (long long)i * jc[i - 1] % mod,
    jcn[i] = (long long)inv[i] * jcn[i - 1] % mod;
  for (long long i = (2), _end_ = (n); i <= _end_; i++)
    pre[i] = a[i - 1] == a[i] ? pre[i - 1] : i - 1;
}
long long C(long long n, long long m) {
  if (n < m) return 0;
  return (long long)jc[n] * jcn[n - m] % mod * jcn[m] % mod;
}
long long ans;
inline void doing() {
  for (long long i = (1), _end_ = (n); i <= _end_; i++) {
    if (a[i] < a[n]) {
      ans = (ans + (long long)a[i] * jc[n] % mod * inv[n - pre[i]] % mod) % mod;
    }
  }
  printf("%lld\n", ans);
}
int main() {
  init();
  doing();
  return 0;
}
