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 double g = 10.0, eps = 1e-12;
const int N = 2000000 + 10, maxn = 1000000 + 10, inf = 0x3f3f3f3f,
          INF = 0x3f3f3f3f3f3f3f3f;
long long f[N], a[maxn];
void init() {
  f[0] = 1;
  for (long long i = 1; i < N; i++) f[i] = f[i - 1] * i % 1000000007;
}
long long quick(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 1000000007;
    a = a * a % 1000000007;
    b >>= 1;
  }
  return ans;
}
long long solve(long long a, long long b) {
  return f[a + b + 1] * quick(b + 1, 1000000007 - 2) % 1000000007;
}
int main() {
  init();
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  sort(a, a + n);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    long long pre = lower_bound(a, a + n, a[i]) - a;
    long long suf = n - pre - 1;
    if (a[i] == a[n - 1]) continue;
    ans = (ans + solve(pre, suf) % 1000000007 * a[i]) % 1000000007;
  }
  printf("%lld\n", ans);
  return 0;
}
