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 long long mod = 1000000007;
const int INF = 0x3f3f3f3f;
const long long INF64 = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-7;
template <class T>
T gcd(T a, T b) {
  if (!b) return a;
  return gcd(b, a % b);
}
const int maxn = 1e6 + 10;
int n;
long long a[maxn], fac, mx = -1;
long long q_pow(long long x, long long k) {
  long long res = 1;
  while (k) {
    if (k & 1) res = res * x % mod;
    k >>= 1;
    x = x * x % mod;
  }
  return res;
}
long long inv(long long x) { return q_pow(x, mod - 2); }
int main() {
  scanf("%d", &n);
  fac = 1;
  for (int i = 2; i <= n; i++) fac = fac * i % mod;
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &a[i]);
    mx = max(mx, a[i]);
  }
  sort(a, a + n);
  long long ans = 0;
  for (int i = 0, le = 0; i < n; i++) {
    if (a[i] == mx) break;
    while (a[le] < a[i]) le++;
    ans = (ans + (a[i] * fac) % mod * inv(n - le) % mod) % mod;
  }
  cout << ans << endl;
  return 0;
}
