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 maxn = 1e6 + 5, mod = 1e9 + 7, inf = 0x3f3f3f3f;
const double eps = 1e-10;
int n, t, a[maxn], rk[maxn];
long long fac[maxn];
long long qpow(long long a, long long b) {
  long long ans = 1, base = a;
  while (b) {
    if (b % 2) {
      ans = ans * base % mod;
    }
    base = base * base % mod;
    b = b >> 1;
  }
  return ans;
}
void init() {
  fac[0] = 1;
  for (int i = 1; i <= 1e6; i++) {
    fac[i] = fac[i - 1] * i % mod;
  }
}
int main() {
  init();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; i++) {
    if (a[i] > a[i - 1]) {
      rk[i] = i - 1;
    } else {
      rk[i] = rk[i - 1];
    }
  }
  long long sum = 0;
  for (int i = 1; i <= n; i++) {
    if (rk[i] == rk[n]) break;
    sum = (sum + 1ll * a[i] * fac[n] % mod * qpow(n - rk[i], mod - 2) % mod) %
          mod;
  }
  printf("%lld\n", sum);
  return 0;
}
