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 const long long Get_Int() {
  long long num = 0, bj = 1;
  char x = getchar();
  while (!isdigit(x)) {
    if (x == '-') bj = -1;
    x = getchar();
  }
  while (isdigit(x)) {
    num = num * 10 + x - '0';
    x = getchar();
  }
  return num * bj;
}
const int maxn = 1000000;
const long long mod = 1e9 + 7;
long long n, fac[maxn + 5], inv[maxn + 5], invf[maxn + 5], a[maxn + 5],
    tmp[maxn + 5], last = 0, sum = 0, ans = 0;
long long C(long long n, long long m) {
  if (!m) return 1;
  return fac[n] * invf[m] % mod * invf[n - m] % mod;
}
int main() {
  fac[0] = 1;
  for (int i = 1; i <= maxn; i++) fac[i] = fac[i - 1] * i % mod;
  inv[1] = invf[1] = invf[0] = 1;
  for (int i = 2; i <= maxn; i++) {
    inv[i] = (mod - (mod / i)) * inv[mod % i] % mod;
    invf[i] = invf[i - 1] * inv[i] % mod;
  }
  n = Get_Int();
  for (int i = 1; i <= n; i++) a[i] = Get_Int();
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    while (a[i] == a[i + 1]) sum++, i++;
    sum++;
    if (i == n) break;
    ans = (ans + (sum - last) * a[i] % mod * C(n, last) % mod * fac[last] %
                     mod * fac[n - last - 1] % mod) %
          mod;
    last = sum;
  }
  printf("%I64d\n", ans);
  return 0;
}
