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 \}.
import java.util.*;

public class Meme
{
    public static void main(String args[])
    {
        Scanner sc= new Scanner(System.in);
        int n=sc.nextInt();
        ArrayList<ArrayList<Integer>> me=new ArrayList<ArrayList<Integer>>();
        me.add(new ArrayList<Integer>());
        for(int x=0;x<n;x++)
        {
            int j=sc.nextInt();
            boolean r=true;
            for(int p=0;p<me.size();p++)
            {
                if(!me.get(p).contains(j))
                {
                    me.get(p).add(j);r=false;break;
                }
            }
            if(r)
            {
                me.add(new ArrayList<Integer>());me.get(me.size()-1).add(j);
            }
        }
        int k=0;
        for(int t=0;t<me.size();t++)
        {
            k+=fact(me.get(t).size())/(fact(t+1)*fact(me.get(t).size()-t-1));
        }
        System.out.println(k);
    }
    public static int fact(int n)
    {
      if (n <= 1)
         return 1;
      else
         return n * fact(n - 1);
   }
}