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() {
  char ch = getchar();
  int x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long fac[1000010], inv[1000010];
int n;
inline void init() {
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) {
    fac[i] = fac[i - 1] * i % 1000000007;
  }
  inv[0] = 0;
  inv[1] = 1;
  for (int i = 2; i <= n; ++i) {
    inv[i] =
        1ll * (1000000007 - 1000000007 / i) * inv[1000000007 % i] % 1000000007;
  }
  for (int i = 2; i <= n; ++i) {
    inv[i] = inv[i - 1] * inv[i] % 1000000007;
  }
}
int a[1000010];
int main() {
  n = read();
  init();
  for (int i = 1; i <= n; ++i) a[i] = read();
  sort(a + 1, a + n + 1);
  int lst = 0;
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i] != a[i - 1]) lst = i - 1;
    if (a[i] != a[n]) {
      ans += fac[n - lst - 1] * fac[n] % 1000000007 * inv[n - lst] %
             1000000007 * a[i] % 1000000007;
      ans %= 1000000007;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}
