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 <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
int arr[1000009], inv[1000009], fac[1000009];
int mod(long long x) { return (x % 1000000007); }
int Fe(int x, int y) {
  if (!y) return 1;
  int h = Fe(x, y / 2);
  h = mod(h * 1LL * h);
  if (y & 1) h = mod(h * 1LL * x);
  return h;
}
int main() {
  fac[0] = inv[0] = 1;
  for (int i = 1; i < 1000009; i++) {
    fac[i] = mod(fac[i - 1] * 1LL * i);
    inv[i] = Fe(i, 1000000007 - 2);
  }
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", arr + i);
  sort(arr + 1, arr + n + 1);
  int l = 0, ans = 0;
  for (int i = 1; arr[i] != arr[n] and i <= n; i++) {
    while (arr[l + 1] < arr[i]) l++;
    ans = mod(ans + mod(mod(arr[i] * 1LL * fac[n]) * 1LL * inv[n - l]));
  }
  printf("%d\n", ans);
  return 0;
}
