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;
mt19937 Rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
template <typename T>
inline void chkmax(T &x, T y) {
  if (x < y) x = y;
}
template <typename T>
inline void chkmin(T &x, T y) {
  if (x > y) x = y;
}
inline int read() {
  int x = 0;
  char c = getchar();
  while (c < 48) c = getchar();
  while (c > 47) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return x;
}
const int maxn = 1e6 + 10, P = 1e9 + 7;
int n, a[maxn], inv[maxn];
int qp(int a, int k) {
  int res = 1;
  for (; k; k >>= 1, a = 1ll * a * a % P) {
    if (k & 1) res = 1ll * res * a % P;
  }
  return res;
}
int main() {
  cin >> n;
  for (int i = (1), iend = (n); i <= iend; ++i) a[i] = read();
  inv[1] = 1;
  int fac = 1;
  for (int i = (2), iend = (n); i <= iend; ++i) {
    fac = 1ll * fac * i % P;
    inv[i] = 1ll * (P - P / i) * inv[P % i] % P;
  }
  sort(a + 1, a + n + 1);
  int res = 0;
  for (int i = (1), iend = (n); i <= iend; ++i) {
    int j = i;
    while (j < n && a[i] == a[j + 1]) j++;
    if (j < n) res = (res + 1ll * a[i] * (j - i + 1) % P * inv[n - i + 1]) % P;
    i = j;
  }
  cout << 1ll * res * fac % P;
  return 0;
}
