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 main() {
  int n;
  cin >> n;
  map<int, int> fcount;
  for (int i = 0; i < n; i++) {
    int d;
    cin >> d;
    fcount[d]++;
  }
  int sz = fcount.size();
  vector<int> counts(sz, 0);
  int max_count = 0;
  int i = 0;
  for (auto it = fcount.begin(); it != fcount.end(); it++) {
    counts[i++] = it->second;
    max_count = max(max_count, it->second);
  }
  vector<long long> width(max_count + 4, 0), inc(max_count + 4, 0);
  width[0] = 1;
  for (int i = 0; i < sz; i++) {
    int cnt = counts[i];
    int layer = 1;
    while (width[layer - 1] != 0) {
      for (int j = layer; j != layer + cnt; j++) {
        inc[j] += width[layer - 1];
      }
      layer++;
    }
    for (int j = 0; j != max_count + 4; j++) {
      width[j] += inc[j];
    }
    inc.assign(max_count + 4, 0);
  }
  cout << width[max_count + 3] % 998244353 << endl;
  return 0;
}
