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(3)
#pragma GCC optimize(2)
#pragma GCC optimize("Ofast")
using namespace std;
const long long mod = 1e9 + 7;
long long fac[1000011];
long long inv(long long p) {
  long long ret = 1, base = p, pw = mod - 2;
  while (pw > 0) {
    if (pw % 2 == 1) {
      ret = ret * base;
      ret %= mod;
      pw--;
    } else {
      base = base * base;
      base %= mod;
      pw /= 2;
    }
  }
  return ret;
}
long long C(int N, int K) {
  return (((fac[N] * inv(fac[K])) % mod) * inv(fac[N - K])) % mod;
}
map<int, int> mp;
int main() {
  int i, n;
  long long ans = 0;
  fac[0] = 1;
  scanf("%d", &n);
  for (i = 1; i < 1000011; ++i) fac[i] = fac[i - 1] * i, fac[i] %= mod;
  for (i = 0; i < n; ++i) {
    int v;
    scanf("%d", &v);
    mp[v]++;
  }
  int cnt = 0;
  mp.erase(mp.rbegin()->first);
  for (auto iter : mp) {
    ans += (((C(n, cnt) * fac[n - cnt - 1]) % mod) * fac[cnt]) % mod *
           (iter.second * iter.first) % mod;
    ans %= mod;
    cnt += iter.second;
  }
  cout << ans << endl;
  return 0;
}
