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 int read() {
  int x = 0, flag = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') flag = 0;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - 48;
    c = getchar();
  }
  return flag ? x : -x;
}
const int N = 1e6 + 7;
const int Mod = 1e9 + 7;
int qpow(long long x, int y = Mod - 2) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = ret * x % Mod;
    x = x * x % Mod, y >>= 1;
  }
  return ret;
}
long long fac[N], a[N], inv[N];
long long C(int x, int y) {
  return x < y ? 0 : fac[x] * inv[y] % Mod * inv[x - y] % Mod;
}
int main() {
  int n = read();
  fac[0] = 1;
  for (int i = 1; i <= n; i++) a[i] = read(), fac[i] = fac[i - 1] * i % Mod;
  inv[n] = qpow(fac[n]);
  for (int i = n - 1; ~i; i--) inv[i] = inv[i + 1] * (i + 1) % Mod;
  sort(a + 1, a + 1 + n);
  int mx = a[n];
  long long ans = 0, q = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == mx) break;
    if (a[i] != a[i - 1]) q = fac[i - 1] * fac[n - i] % Mod * C(n, i - 1) % Mod;
    ans = (ans + a[i] * q % Mod) % Mod;
  }
  printf("%lld", ans);
}
