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;
namespace io {
const int __SIZE = (1 << 21) + 1;
char ibuf[__SIZE], *iS, *iT, obuf[__SIZE], *oS = obuf, *oT = oS + __SIZE - 1,
                                           __c, qu[55];
int __f, qr, _eof;
inline void flush() { fwrite(obuf, 1, oS - obuf, stdout), oS = obuf; }
inline void gc(char &x) {
  x = (iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, __SIZE, stdin),
                   (iS == iT ? EOF : *iS++))
                : *iS++);
}
inline void pc(char x) {
  *oS++ = x;
  if (oS == oT) flush();
}
inline void pstr(const char *s) {
  int __len = strlen(s);
  for (__f = 0; __f < __len; ++__f) pc(s[__f]);
}
inline void gstr(char *s) {
  for (__c = (iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, __SIZE, stdin),
                         (iS == iT ? EOF : *iS++))
                       : *iS++);
       __c < 32 || __c > 126 || __c == ' ';)
    __c = (iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, __SIZE, stdin),
                       (iS == iT ? EOF : *iS++))
                    : *iS++);
  for (; __c > 31 && __c < 127 && __c != ' ';
       ++s, __c = (iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, __SIZE, stdin),
                              (iS == iT ? EOF : *iS++))
                            : *iS++))
    *s = __c;
  *s = 0;
}
template <class I>
inline bool gi(I &x) {
  _eof = 0;
  for (__f = 1,
      __c = (iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, __SIZE, stdin),
                        (iS == iT ? EOF : *iS++))
                      : *iS++);
       (__c < '0' || __c > '9') && !_eof;
       __c = (iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, __SIZE, stdin),
                         (iS == iT ? EOF : *iS++))
                       : *iS++)) {
    if (__c == '-') __f = -1;
    _eof |= __c == EOF;
  }
  for (x = 0; __c <= '9' && __c >= '0' && !_eof;
       __c = (iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, __SIZE, stdin),
                         (iS == iT ? EOF : *iS++))
                       : *iS++))
    x = x * 10 + (__c & 15), _eof |= __c == EOF;
  x *= __f;
  return !_eof;
}
template <class I>
inline void print(I x) {
  if (!x) pc('0');
  if (x < 0) pc('-'), x = -x;
  while (x) qu[++qr] = x % 10 + '0', x /= 10;
  while (qr) pc(qu[qr--]);
}
struct Flusher_ {
  ~Flusher_() { flush(); }
} io_flusher_;
}  // namespace io
using io::gc;
using io::gi;
using io::gstr;
using io::pc;
using io::print;
using io::pstr;
using ll = long long;
using pii = pair<int, int>;
const int MOD = 1000000007;
int inv[1000006];
int nf = 1;
int n;
int ar[1000006];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  inv[0] = inv[1] = 1;
  gi(n);
  for (int i = (2); i < (n + 2); i++)
    inv[i] = 1ll * (MOD - MOD / i) * inv[MOD % i] % MOD;
  for (int i = (1); i < ((n) + 1); i++) nf = 1ll * nf * i % MOD;
  for (int i = (0); i < (n); i++) gi(ar[i]);
  sort(ar, ar + n);
  int t = ar[n - 1], N = n;
  while (n && ar[n - 1] == t) n--;
  int cnt = 0, ans = 0;
  for (int i = (0); i < (n); i++) {
    while (ar[cnt] < ar[i]) cnt++;
    ans = (ans + 1ll * ar[i] * nf % MOD * inv[N - cnt]) % MOD;
  }
  print(ans);
}
