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;
const int N = 2e5 + 7;
int a[N], f[N];
const int MOD = 998244353;
int mul(int a, int b) { return (long long)a * b % MOD; }
int add(int a, int b) {
  a %= MOD;
  b %= MOD;
  return (a + b) % MOD;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  clock_t ttt = clock();
  int n;
  cin >> n;
  unordered_map<int, int> b;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    b[a[i]]++;
  }
  if (n == 1) {
    cout << 1 << '\n';
    return 0;
  }
  int k = b.size();
  f[0] = f[1] = 1;
  for (int i = 2; i <= k; ++i) {
    f[i] = mul(f[i - 1], i);
  }
  int ans = 0;
  for (auto &v : b) {
    if (v.second > 1) ans = add(ans, 1);
  }
  for (int i = 1; i < k; ++i) {
    ans = add(ans, f[i]);
  }
  cout << ans << '\n';
  cerr << "Time taken " << (double)(clock() - ttt) / CLOCKS_PER_SEC
       << " seconds\n";
  return 0;
}
