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;
inline long long read() {
  long long f = 1, x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f = (ch == '-') ? -1 : 1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return f * x;
}
const int N = 1e6 + 5;
const int P = 1e9 + 7;
int fac[N], inv[N];
int add(int a, int b) {
  a += b;
  return a >= P ? a - P : a;
}
int mul(int a, int b) { return (long long)a * b % P; }
int sub(int a, int b) {
  a -= b;
  return a < 0 ? a + P : a;
}
inline int C(int n, int m) { return mul(fac[n], mul(inv[m], inv[n - m])); }
int a[N];
int modexp(int a, int x) {
  int ret = 1;
  while (x) {
    if (x & 1) ret = mul(ret, a);
    a = mul(a, a);
    x >>= 1;
  }
  return ret;
}
int n;
int main() {
  n = read();
  for (int i = (0); i < (n); ++i) a[i] = read();
  fac[0] = 1;
  for (int i = (1); i < (n + 1); ++i) fac[i] = mul(fac[i - 1], i);
  inv[n] = modexp(fac[n], P - 2);
  for (int i = (n)-1; i >= (0); --i) inv[i] = mul(inv[i + 1], i + 1);
  sort(a, a + n);
  int ans = 0;
  for (int i = (0); i < (n); ++i) {
    int l = i, r = i;
    while (a[r] == a[l]) ++r;
    if (r == n) break;
    int L = i;
    ans = add(ans, mul(mul(fac[n - L - 1], mul(fac[L], mul(a[i], C(n, n - L)))),
                       r - l));
    i = r - 1;
  }
  cout << ans << endl;
  return 0;
}
