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>
#pragma GCC target("f16c")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse3", "sse2", "sse")
#pragma GCC optimize(3, "Ofast", "inline")
#pragma GCC optimize("no-stack-protector")
#pragma GCC target("avx", "sse4", "sse4.1", "sse4.2", "ssse3")
using namespace std;
const long long mod = 1e9 + 7;
long long n, ans;
long long ai[1005 * 1005];
long long fac[1005 * 1005], ni[1005 * 1005];
inline long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
inline void Init() {
  fac[0] = fac[1] = 1;
  ni[0] = ni[1] = 1;
  for (long long i = 2; i <= 1e6 + 50; ++i) {
    fac[i] = (1LL * fac[i - 1] * i) % mod;
    ni[i] = (1LL * ni[mod % i] * (mod - mod / i)) % mod;
  }
}
signed main() {
  scanf("%lld", &n);
  Init();
  for (long long i = 1; i <= n; ++i) scanf("%lld", &ai[i]);
  sort(ai + 1, ai + 1 + n);
  for (long long i = 1; i <= n;) {
    long long j = i;
    while (ai[j + 1] == ai[j]) j++;
    if (j == n) break;
    ans =
        ans + (j - i + 1) * (ai[i] * fac[n] % mod * ni[n - i + 1] % mod) % mod;
    i = j + 1;
  }
  printf("%lld", ans % mod);
}
