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, a[1000010];
long long ans, f[1000010], g[1000010], num[1000010], fac[1000010], sum[1000010];
template <class Aqua>
inline void read(Aqua &s) {
  s = 0;
  Aqua f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) s = s * 10 + c - '0', c = getchar();
  s *= f;
}
long long mpow(long long a, long long b) {
  long long s = 1;
  for (; b; b >>= 1, a = a * a % 1000000007)
    if (b & 1) s = s * a % 1000000007;
  return s;
}
long long com(long long a) {
  return fac[n] * mpow(fac[a], 1000000007 - 2) % 1000000007;
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  sort(a + 1, a + 1 + n);
  int cnt = 1;
  num[1] = 1;
  for (int i = 2; i <= n; num[cnt]++, i++)
    if (a[i] != a[i - 1]) cnt++;
  unique(a + 1, a + 1 + n);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = (long long)fac[i - 1] * i % 1000000007;
  long long s = 0;
  for (int i = 1; i <= cnt; i++) f[i] = fac[s], s += num[i];
  s = 0;
  for (int i = cnt; i; i--) sum[i] = (s += num[i]), g[i] = fac[s - 1];
  for (int i = 1; i < cnt; i++)
    (ans += a[i] * g[i] % 1000000007 * num[i] % 1000000007 * com(sum[i]) %
            1000000007) %= 1000000007;
  printf("%I64d\n", ans);
  return 0;
}
