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 int N = 1e6 + 6, mod = 1e9 + 7;
int n,
    mult[N] =
        {
            1,
},
    ans;
int num[N];
inline int inv(int x) {
  int res = 1;
  for (int y = mod - 2; y; y >>= 1, x = 1ll * x * x % mod)
    if (y & 1) res = 1ll * res * x % mod;
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", num + i);
    mult[i] = 1ll * mult[i - 1] * i % mod;
  }
  sort(num + 1, num + 1 + n);
  for (int i = 1; i <= n; i++) {
    if (num[i] == num[n]) break;
    int rk = lower_bound(num + 1, num + 1 + n, num[i]) - num;
    ans = (ans + 1ll * mult[n] * inv(n - rk + 1) % mod * num[i]) % mod;
  }
  printf("%d\n", ans);
}
