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 long long mod = 1e9 + 7;
const long long inf = (1ll << 62);
const int MX = 1e6 + 9;
int n;
long long a[MX], ans, orig[MX], fac[MX], cnt[MX], pre[MX];
map<long long, long long> hashy;
long long POW(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  long long ans = POW(x, y / 2) % mod;
  ans *= ans;
  ans %= mod;
  if (y % 2) ans *= x;
  return ans % mod;
}
long long modv(long long x) {
  long long ret = fac[n];
  long long ret1 = POW(x, mod - 2) % mod;
  return (ret * ret1) % mod;
}
long long C(long long n, long long r) {
  long long ret = fac[n];
  long long ret1 = fac[r];
  ret1 = (fac[n - r] * ret1) % mod;
  ret = POW(ret1, mod - 2) % mod;
  return (ret * ret1) % mod;
}
int main() {
  fac[0] = 1;
  cin >> n;
  for (long long i = 1; i <= n; i++) fac[i] = (fac[i - 1] * i) % mod;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    hashy[a[i]] = 1;
  }
  int tt = 0;
  for (auto pp : hashy) {
    hashy[pp.first] = ++tt;
    orig[tt] = pp.first;
  }
  for (int i = 1; i <= n; i++) {
    cnt[hashy[a[i]]]++;
  }
  for (int i = 1; i <= n; i++) {
    pre[i] = pre[i - 1] + cnt[i];
  }
  for (int i = 1; i <= n; i++) {
    if (i == tt) break;
    long long gr = n - pre[i - 1];
    long long x = (cnt[i] * orig[i]) % mod;
    x = (x * modv(gr)) % mod;
    ans += x;
    ans %= mod;
  }
  cout << ans << endl;
}
