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 char nc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline void Read(int& x) {
  char c = nc();
  for (; c < '0' || c > '9'; c = nc())
    ;
  for (x = 0; c >= '0' && c <= '9'; x = (x << 3) + (x << 1) + c - 48, c = nc())
    ;
}
const int N = 1000010;
const int M = 1000000007;
int k, n, m;
int s, Ans, x;
int inv[N];
int a[N], fac[N], f[N];
int main() {
  Read(n);
  inv[0] = inv[1] = 1;
  for (int i = 2; i <= n; i++) inv[i] = 1ll * inv[M % i] * (M - M / i) % M;
  for (int i = 1; i <= n; i++) Read(a[i]);
  sort(a + 1, a + n + 1);
  for (int i = 1; i < n && a[i] < a[n]; i++) {
    if (a[i] == a[i - 1])
      x++;
    else
      x = 0;
    Ans = (Ans + 1ll * a[i] * inv[n - (i - x - 1)]) % M;
  }
  for (int i = 2; i <= n; i++) Ans = 1ll * Ans * i % M;
  cout << Ans << endl;
  return 0;
}
