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;
template <typename T>
using min_queue = priority_queue<T, vector<T>, greater<T> >;
const long long MOD = 1e9 + 7;
const int MAX = 1000001;
int N;
int a[MAX];
long long fac[MAX];
inline long long pw(long long n, long long k) {
  n %= MOD;
  long long res = 1;
  while (k) {
    if (k & 1) res = (res * n) % MOD;
    n = (n * n) % MOD;
    k >>= 1;
  }
  return res;
}
inline long long inv(long long n) { return pw(n, MOD - 2); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  for (int i = 0; i < N; i++) cin >> a[i];
  fac[0] = 1;
  for (int i = 1; i <= N; i++) fac[i] = (fac[i - 1] * i) % MOD;
  sort(a, a + N);
  long long res = 0;
  int j = 0;
  for (int i = 0; a[i] != a[N - 1]; i++) {
    if (i > 0 && a[i] != a[i - 1]) j = i;
    long long tmp = fac[N] * inv(N - j) % MOD;
    res = (res + (tmp * a[i]) % MOD) % MOD;
  }
  cout << res << '\n';
  return 0;
}
