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 N = 1e6 + 5, mod = 1e9 + 7;
int n, x;
map<int, int> mp;
int f[N], inv[N];
int inverse(int b) {
  int p = mod - 2, ret = 1;
  while (p) {
    if (p & 1) ret = 1ll * ret * b % mod;
    b = 1ll * b * b % mod;
    p >>= 1;
  }
  return ret;
}
int choose(int n, int r) {
  return 1ll * f[n] * inv[r] % mod * inv[n - r] % mod;
}
int main() {
  scanf("%d", &n);
  f[0] = inv[0] = 1;
  for (int i = 1; i <= n; ++i) {
    f[i] = 1ll * i * f[i - 1] % mod;
    inv[i] = inverse(f[i]);
  }
  int mx = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &x);
    mx = max(mx, x);
    ++mp[x];
  }
  vector<pair<int, int>> mpv;
  map<int, int> prv_cnt;
  for (auto go : mp) mpv.push_back(go);
  int cnt = 0;
  for (auto go : mpv) {
    prv_cnt[go.first] = cnt;
    cnt += go.second;
  }
  reverse(mpv.begin(), mpv.end());
  cnt = 0;
  int ans = 0;
  for (auto go : mpv) {
    if (go.first != mx) {
      int cur = 1ll * go.second * f[cnt + go.second - 1] % mod;
      cur = 1ll * cur * f[prv_cnt[go.first]] % mod *
            choose(n, prv_cnt[go.first]) % mod;
      cur = 1ll * cur * go.first % mod;
      ans = (ans + cur) % mod;
    }
    cnt += go.second;
  }
  printf("%d", ans);
  return 0;
}
