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 mod = 1e9 + 7;
const int mx = 1e6 + 10;
long long da[mx + 10];
long long fac[mx + 10];
long long pre[mx + 10];
multiset<long long> s;
long long qpow(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ans;
}
long long inv(int t) { return qpow(t, mod - 2); }
int main() {
  long long n;
  fac[1] = 1 % mod;
  scanf("%lld", &n);
  for (int i = 1; i <= n; ++i) scanf("%lld", &da[i]), s.insert(da[i]);
  for (int i = 2; i < mx; ++i) fac[i] = fac[i - 1] * i % mod;
  sort(da + 1, da + 1 + n);
  int num = unique(da + 1, da + 1 + n) - (da + 1);
  for (int i = 1; i <= num; ++i) pre[i] = pre[i - 1] + s.count(da[i]);
  long long ans = 0;
  for (int i = 1; i < num; ++i) {
    ans = ans + fac[n] * inv(n - pre[i - 1]) % mod * da[i] % mod *
                    s.count(da[i]) % mod;
    ans %= mod;
  }
  printf("%lld\n", ans);
  return 0;
}
