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;
int n, A[1000005], ans;
int add(int a, int b) {
  a += b;
  return a >= mod ? a - mod : a;
}
int sub(int a, int b) {
  a -= b;
  return a < 0 ? a + mod : a;
}
int mul(int a, int b) { return 1LL * a * b % mod; }
int Pow(int a, int b) {
  int r = 1;
  while (b) {
    if (b & 1) r = mul(r, a);
    a = mul(a, a);
    b >>= 1;
  }
  return r;
}
int fact[1000005], finv[1000005];
void init() {
  int i;
  fact[0] = 1;
  for (i = 1; i <= n; i++) fact[i] = mul(fact[i - 1], i);
  finv[n] = Pow(fact[n], mod - 2);
  for (i = n; i >= 1; i--) finv[i - 1] = mul(finv[i], i);
}
int C(int a, int b) { return mul(fact[a], mul(finv[b], finv[a - b])); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
  init();
  sort(A + 1, A + 1 + n);
  int x = 0, y = 0;
  for (int i = 1; i <= n; i++) {
    if (A[i] == A[n]) break;
    y = lower_bound(A + 1, A + 1 + n, A[i]) - A - 1;
    x = n - 1 - y;
    ans = add(ans, mul(mul(fact[y], fact[x]), mul(C(n, x + 1), A[i])));
  }
  printf("%d", ans);
}
