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 gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int read() {
  int x = 0;
  char ch = getchar();
  bool positive = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') positive = 0;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return positive ? x : -x;
}
inline void write(int a) {
  if (a >= 10) write(a / 10);
  putchar('0' + a % 10);
}
inline void writeln(int a) {
  if (a < 0) {
    a = -a;
    putchar('-');
  }
  write(a);
  puts("");
}
const int N = 1000005, mod = 1000000007;
inline int ksm(int a, int b) {
  int t = 1, y = a;
  while (b) {
    if (b & 1) t = (long long)t * y % mod;
    y = (long long)y * y % mod;
    b >>= 1;
  }
  return t;
}
int n, fac[N], a[N], mx, ans, fff[N], ycl[N];
inline int p(int n, int m) { return (long long)fac[n] * fff[n - m] % mod; }
int main() {
  n = read();
  fff[0] = fac[0] = 1;
  for (int i = 1; i <= n + 1; i++) {
    if (i <= n) mx = max(mx, a[i] = read());
    fac[i] = (long long)fac[i - 1] * i % mod;
    fff[i] = ksm(fac[i], mod - 2);
    ycl[i] = ((long long)ycl[i - 1] * i + 1) % mod;
  }
  sort(&a[1], &a[n + 1]);
  for (int i = 1; i <= n; i++)
    if (a[i] != mx) {
      int t = lower_bound(&a[1], &a[n + 1], a[i]) - a - 1;
      ans = (ans + (long long)a[i] * fac[n] % mod *
                       ksm((long long)(n - t) * fac[t] % mod, mod - 2) % mod *
                       fac[t]) %
            mod;
    }
  cout << ans << endl;
}
