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;
template <typename TH>
void _dbg(const char* sdbg, TH h) {
  cerr << sdbg << " = " << h << "\n";
}
template <typename TH, typename... TA>
void _dbg(const char* sdbg, TH h, TA... t) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << " = " << h << ",";
  _dbg(sdbg + 1, t...);
}
const int N = 1e6 + 6, MOD = 1e9 + 7;
int n, a[N];
int fact[N], invFact[N];
int Pow(int n, int k) {
  int res = 1;
  while (k > 0) {
    if (k & 1) {
      res = (int64_t)res * n % MOD;
    }
    n = (int64_t)n * n % MOD;
    k >>= 1;
  }
  return res;
}
int C(int n, int k) {
  if (k > n) {
    return 0;
  }
  int res = fact[n];
  res = (int64_t)res * invFact[k] % MOD;
  res = (int64_t)res * invFact[n - k] % MOD;
  return res;
}
void Brute() {
  vector<int> p;
  for (int i = 1; i <= n; ++i) {
    p.push_back(i);
  }
  int ans = 0;
  map<int, int> cnt;
  do {
    int m = 1, f = 0;
    bool flag = false;
    for (int i = 2; i <= n; ++i) {
      if (a[p[i - 1]] > a[p[m - 1]]) {
        f += a[p[m - 1]];
        cnt[p[m - 1]]++;
        if (p[m - 1] == 1) {
          flag = true;
        }
        m = i;
      }
    }
    if (flag) {
      for (int j = 1; j <= n; ++j) {
        cout << a[p[j - 1]] << ' ';
      }
      cout << '\n';
    }
    ans += f;
  } while (next_permutation(begin(p), end(p)));
  cout << ans << '\n';
  for (auto it : cnt) {
    _dbg("it.first, it.second", it.first, it.second);
  }
  exit(0);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  map<int, int> freq;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    freq[a[i]]++;
  }
  fact[0] = 1;
  for (int i = 1; i <= n; ++i) {
    fact[i] = (int64_t)fact[i - 1] * i % MOD;
  }
  invFact[n] = Pow(fact[n], MOD - 2);
  for (int i = n - 1; i >= 0; --i) {
    invFact[i] = (int64_t)invFact[i + 1] * (i + 1) % MOD;
  }
  int ans = 0, x = 0;
  for (auto it : freq) {
    int y = n - x - it.second;
    if (y > 0) {
      int cnt = fact[x];
      cnt = (int64_t)cnt * fact[n - x - 1] % MOD;
      cnt = (int64_t)cnt * C(n, n - x) % MOD;
      int each = (int64_t)cnt * it.first % MOD;
      ans = (ans + (int64_t)each * it.second % MOD) % MOD;
    }
    x += it.second;
  }
  cout << ans;
  return 0;
}
