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 V = vector<t>;
using ll = long long;
constexpr int mod = 1e9 + 7;
int n;
V<int> tb;
V<int> sil;
inline int fastpow(int p, int w) {
  int res = 1;
  while (w > 0) {
    if (w & 1) res = ((ll)res * p) % mod;
    p = ((ll)p * p) % mod;
    w >>= 1;
  }
  return res;
}
inline int binomial(int gor, int dol) {
  int tmp = ((ll)sil[dol] * sil[gor - dol]) % mod;
  tmp = fastpow(tmp, mod - 2);
  gor = sil[gor];
  return ((ll)gor * tmp) % mod;
}
inline int licz(int a, int b, int c) {
  int res = ((ll)sil[a] * sil[b]) % mod;
  res = ((ll)res * sil[c]) % mod;
  res = ((ll)res * binomial(b + c - 1, c - 1)) % mod;
  res = ((ll)res * binomial(n, a)) % mod;
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  tb.resize(n);
  sil.resize(n + 1);
  for (int i = 0; i < n; ++i) cin >> tb[i];
  sil[0] = 1;
  for (int i = 1; i <= n; ++i) sil[i] = ((ll)sil[i - 1] * i) % mod;
  sort(begin(tb), end(tb));
  int res = 0;
  int pop = -1;
  for (int i = 0; i < n - 1; ++i)
    if (tb[i + 1] != tb[i]) {
      int tmp = licz(pop + 1, n - i - 1, i - pop);
      res = ((ll)res + (ll)tmp * tb[i]) % mod;
      pop = i;
    }
  cout << res << '\n';
}
