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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
const int maxn = 1001000;
const int mod = 1000000007;
int a[maxn], f[maxn], r[maxn];
int main() {
  int n;
  scanf("%d", &n);
  for (int e = 1; e <= n; e++) scanf("%d", a + e);
  auto mul = [](int x, int y) { return 1ll * x * y % mod; };
  f[0] = 1;
  for (int e = 1; e <= n; e++) f[e] = mul(e, f[e - 1]);
  r[n + 1] = 1;
  for (int e = n; e >= 1; e--) r[e] = mul(e, r[e + 1]);
  sort(a + 1, a + 1 + n);
  int small = 0;
  int i = 1, j = 1;
  int ans = 0;
  while (i <= n) {
    while (j <= n && a[i] == a[j]) j++;
    if (j == n + 1) break;
    ans += mul(j - i, mul(a[i], mul(f[n - small - 1], r[n - small + 1])));
    small += j - i;
    if (ans >= mod) ans -= mod;
    i = j;
  }
  cout << ans << endl;
  return 0;
}
