You are given an integer x represented as a product of n its prime divisors p_1 ⋅ p_2, ⋅ … ⋅ p_n. Let S be the set of all positive integer divisors of x (including 1 and x itself).

We call a set of integers D good if (and only if) there is no pair a ∈ D, b ∈ D such that a ≠ b and a divides b.

Find a good subset of S with maximum possible size. Since the answer can be large, print the size of the subset modulo 998244353.

Input

The first line contains the single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of prime divisors in representation of x.

The second line contains n integers p_1, p_2, ..., p_n (2 ≤ p_i ≤ 3 ⋅ 10^6) — the prime factorization of x.

Output

Print the maximum possible size of a good subset modulo 998244353.

Examples

Input


3
2999999 43 2999957


Output


3


Input


6
2 3 2 3 2 2


Output


3

Note

In the first sample, x = 2999999 ⋅ 43 ⋅ 2999957 and one of the maximum good subsets is \{ 43, 2999957, 2999999 \}.

In the second sample, x = 2 ⋅ 3 ⋅ 2 ⋅ 3 ⋅ 2 ⋅ 2 = 144 and one of the maximum good subsets is \{ 9, 12, 16 \}.
#include <bits/stdc++.h>
using namespace std;
int p[200005], n, m;
long long a[200005], ans;
inline int read() {
  int x = 0;
  char ch = getchar();
  while (ch > '9' || ch < '0') ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  return x;
}
int main() {
  n = read();
  if (n == 1) {
    printf("1\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) p[i] = read();
  sort(p + 1, p + n + 1);
  m = 1;
  a[1] = 1;
  for (int i = 2; i <= n; i++)
    if (p[i] == p[m])
      a[m]++;
    else
      p[++m] = p[i], a[m] = 1;
  sort(a + 1, a + m + 1);
  ans = m;
  for (int i = 1; i < m; i++)
    ans = (ans + (a[i] - 1) * (m - i) % 998244353) % 998244353;
  printf("%lld\n", ans % 998244353);
  return 0;
}
