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 mod = 1e9 + 7;
const int N = 1000010;
inline void read(int &wh) {
  wh = 0;
  int f = 1;
  char w = getchar();
  while (w < '0' || w > '9') {
    if (w == '-') f = -1;
    w = getchar();
  }
  while (w <= '9' && w >= '0') {
    wh = wh * 10 + w - '0';
    w = getchar();
  }
  wh *= f;
  return;
}
inline int mul(int s1, int s2) { return (long long)s1 * s2 % mod; }
inline void add(int &s1, int s2) {
  s1 = s1 + s2 >= mod ? s1 + s2 - mod : s1 + s2;
}
int m, a[N];
int jc[N], fjc[N];
inline int pow(int s1, int s2) {
  if (s2 == 0) return 1;
  if (s2 == 1) return s1;
  int an = pow(s1, s2 >> 1);
  if (s2 & 1)
    return mul(mul(an, an), s1);
  else
    return mul(an, an);
}
void init_c() {
  jc[0] = 1;
  for (int i = 1; i <= m; i++) {
    jc[i] = mul(jc[i - 1], i);
  }
  fjc[m] = pow(jc[m], mod - 2);
  for (int i = m - 1; i >= 0; i--) {
    fjc[i] = mul(fjc[i + 1], i + 1);
  }
  return;
}
inline int c(int s1, int s2) {
  return s1 < s2 ? 0 : mul(mul(jc[s1], fjc[s1 - s2]), fjc[s2]);
}
signed main() {
  read(m);
  for (int i = 1; i <= m; i++) read(a[i]);
  init_c();
  sort(a + 1, a + m + 1);
  int ans = 0, now = 0;
  for (int i = 1; i <= m; i++) {
    if (a[i] == a[m]) break;
    if (a[i] != a[i - 1]) now = mul(mul(jc[i - 1], jc[m - i]), c(m, i - 1));
    add(ans, mul(now, a[i]));
  }
  printf("%d", ans);
  return 0;
}
