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 INF = (int)1e9;
const int mod = (int)1e9 + 7;
const int MAX = (int)1e6 + 5;
long long val[MAX], re_val[MAX];
inline long long mypow(long long v, int t, int md) {
  long long ret = 1;
  while (t) {
    if (t & 1) ret = ret * v % md;
    v = v * v % md;
    t >>= 1;
  }
  return ret;
}
void build() {
  long long cur = val[0] = 1;
  for (int i = (1); i < (MAX); ++i) {
    cur = cur * i % mod;
    val[i] = cur;
  }
  for (int i = 0; i < (MAX); ++i) re_val[i] = mypow(val[i], mod - 2, mod);
  return;
}
long long store[MAX];
long long n;
inline long long sol(int p) {
  long long ret = n * val[n - 1] % mod;
  ret = ret * mypow(n - p + 1, mod - 2, mod) % mod;
  return ret;
}
map<long long, int> num;
int main(void) {
  build();
  scanf("%lld", &n);
  for (int i = (1); i < (n + 1); ++i) {
    scanf("%lld", store + i);
    num[store[i]]++;
  }
  sort(store + 1, store + n + 1);
  long long trace = n;
  long long d = store[trace];
  while (store[trace] == d) trace--;
  long long res = 0;
  for (int i = (1); i < (trace + 1); ++i) {
    if (i != 1 && store[i] == store[i - 1]) continue;
    res = (res + (sol(i) * store[i] % mod) * num[store[i]]) % mod;
  }
  printf("%lld", res);
  return 0;
}
