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[200000 + 5];
int n;
struct Node {
  int p;
  int num;
  friend bool operator<(Node p, Node q) { return p.num < q.num; }
} a[200000 + 5];
int f[200000 + 5][2];
int mx(int a, int b) { return a > b ? a : b; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
  }
  sort(p + 1, p + 1 + n);
  int len = 0;
  for (int i = 1, j; i <= n; i = j + 1) {
    j = i;
    while (j <= n && p[j] == p[i]) {
      j++;
    }
    j--;
    a[++len].num = j - i + 1;
    a[len].p = p[i];
  }
  sort(a + 1, a + 1 + len);
  for (int i = 1; i <= len; i++) {
    f[i][0] = mx(f[i - 1][0], f[i - 1][1]) + 1;
    if (i > 1) {
      f[i][1] = f[i - 1][0] + a[i - 1].num;
    }
  }
  printf("%d\n", mx(f[len][0], f[len][1]));
  return 0;
}
