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;
const long long MOD = 1e9 + 7;
const long long INF = 1e18 + 7;
const long long MAXN = 1e6 + 7;
long long f[MAXN], fac, n, m, ans;
long long qp(long long x, long long y) {
  long long res = 1ll;
  while (y) {
    if (y & 1ll) res = res * x % MOD;
    y >>= 1ll;
    x = x * x % MOD;
  }
  return (res % MOD + MOD) % MOD;
}
signed main() {
  long long i, j, k, l, ca = 0;
  scanf("%lld", &n);
  for (fac = i = 1ll; i <= n; i++) {
    scanf("%lld", &f[i]);
    fac = fac * i % MOD;
  }
  sort(f + 1, f + n + 1);
  f[n + 1] = f[f[0] = n];
  while (f[n] == f[n + 1]) n--;
  for (i = 1; i <= n; i++) {
    if (f[i] != f[i - 1])
      m = f[i] * qp(f[0] - i + 1ll, MOD - 2) % MOD * fac % MOD;
    (ans += m) %= MOD;
  }
  printf("%lld\n", (ans % MOD + MOD) % MOD);
  return 0;
}
