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;
long long mod = 1e9 + 7;
const int maxn = 1e6 + 5;
int n, m, t;
long long fac[maxn];
long long sum[maxn];
long long a[maxn];
long long ans = 0;
int read() {
  int x;
  scanf("%d", &x);
  return x;
}
long long qpow(long long b, long long x) {
  long long ret = 1;
  while (x) {
    if (x & 1) {
      ret = ret * b % mod;
    }
    b = b * b % mod;
    x >>= 1;
  }
  return ret;
}
int init() {
  long long i, j;
  fac[1] = 1;
  for (i = 2; i <= n; i++) {
    fac[i] = fac[i - 1] * i % mod;
  }
  return 0;
}
int main() {
  int i, j;
  int l, x;
  n = read();
  init();
  for (i = 1; i <= n; i++) {
    a[i] = read();
  }
  sort(a + 1, a + n + 1);
  x = n;
  while (a[x] == a[x - 1]) {
    x--;
  }
  for (i = 1; i < x; i++) {
    if (a[i] != a[i - 1]) {
      l = i;
    }
    ans = (ans + a[i] * fac[n] % mod * qpow(n - l + 1, mod - 2)) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}
