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>
const long long P = 1e9 + 7;
const int N = 2e6 + 5;
long long a[N], Fac[N], FacInv[N], ans = 0;
int n;
template <typename T>
T qpow(T x, T y) {
  T res = 1;
  for (; y; y >>= 1, x = x * x % P)
    if (y & 1) res = (res * x) % P;
  return res;
}
void pre(int MAXN) {
  Fac[0] = 1;
  for (int i = 1; i <= MAXN; i++) Fac[i] = Fac[i - 1] * i % P;
  FacInv[MAXN] = qpow(Fac[MAXN], P - 2);
  for (int i = MAXN - 1; ~i; i--) FacInv[i] = FacInv[i + 1] * (i + 1) % P;
}
long long C(long long n, long long m) {
  return Fac[n] * FacInv[m] % P * FacInv[n - m] % P;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  pre(n + 1);
  std::sort(a + 1, a + 1 + n);
  long long sum = 0;
  for (long long i = 1; i < n; i++) {
    ++sum;
    if (i != n && a[i] == a[i + 1]) continue;
    long long k = i - sum;
    ans = (ans +
           sum * a[i] % P * Fac[k] % P * Fac[n - 1 - k] % P * C(n, k) % P + P) %
          P;
    sum = 0;
  }
  printf("%lld\n", ans);
  return 0;
}
