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 INF = 1e18, MAXN = 1e6 + 100, MOD = 1e9 + 7;
long long a[MAXN], fact[MAXN], ans;
long long pw(long long a, long long b) {
  if (!b) return 1;
  long long x = pw(a, b / 2);
  x = x * x % MOD;
  if (b % 2) x = x * a % MOD;
  return x;
}
long long c(long long n, long long r) {
  long long res = fact[n];
  res = res * (pw(fact[r], MOD - 2)) % MOD;
  res = res * (pw(fact[n - r], MOD - 2)) % MOD;
  return res;
}
void fix(long long &x) {
  x %= MOD;
  if (x < 0) {
    long long www = -x / MOD;
    x += www * MOD;
  }
  while (x < 0) x += MOD;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  fact[0] = 1;
  for (long long i = 1; i < MAXN; i++) fact[i] = fact[i - 1] * i % MOD;
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  long long kochiktar = 0;
  for (long long i = n - 1; i >= 0; i--) {
    long long j = 1;
    while (i > 0 && a[i] == a[i - 1]) i--, j++;
    if (kochiktar == 0) {
      kochiktar += j;
      continue;
    }
    long long x = j * a[i] *
                  (((c(n, kochiktar + j) * fact[kochiktar + j - 1] % MOD) *
                    fact[n - kochiktar - j]) %
                   MOD);
    ans += x;
    fix(ans);
    kochiktar += j;
  }
  cout << ans << endl;
}
