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;
void prn(vector<long long unsigned> v) {
  for (long long unsigned i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << endl;
}
const long long unsigned mod = 998244353;
const long long unsigned root = 72390235;
const long long unsigned root_1 = 967518384;
const long long unsigned root_pw = 1 << 23;
long long unsigned power(long long unsigned b, long long unsigned p) {
  if (p == 0) return 1;
  if (p == 1) return b;
  long long unsigned x = power(b, p / 2) % mod;
  x = (x * x) % mod;
  if (p & 1) x = (x * b % mod) % mod;
  return x;
}
long long unsigned inverse(long long unsigned n, long long unsigned m) {
  long long unsigned x = power(n, m - 2);
  return x % mod;
}
void fft(vector<long long unsigned>& a, bool invert) {
  long long unsigned n = a.size();
  for (long long unsigned i = 1, j = 0; i < n; i++) {
    long long unsigned bit = n >> 1;
    for (; j & bit; bit >>= 1) j ^= bit;
    j ^= bit;
    if (i < j) swap(a[i], a[j]);
  }
  for (long long unsigned len = 2; len <= n; len <<= 1) {
    long long unsigned wlen = invert ? root_1 : root;
    for (long long unsigned i = len; i < root_pw; i <<= 1)
      wlen = (long long unsigned)((1LLU * wlen * wlen) % mod);
    for (long long unsigned i = 0; i < n; i += len) {
      long long unsigned w = 1;
      for (long long unsigned j = 0; j < len / 2; j++) {
        long long unsigned u = a[i + j],
                           v = (long long unsigned)((1LL * a[i + j + len / 2] *
                                                     w) %
                                                    mod);
        a[i + j] = u + v < mod ? u + v : u + v - mod;
        a[i + j + len / 2] = u - v >= 0 ? u - v : u - v + mod;
        if (a[i + j] > mod or a[i + j + len / 2] > mod) {
          cout << i << " " << j << endl;
          a[i + j] = a[i + j] % mod;
          a[i + j + len / 2] = a[i + j + len / 2] % mod;
        }
        w = (long long unsigned)((1LL * w * wlen) % mod);
      }
    }
  }
  if (invert) {
    long long unsigned n_1 = inverse(n, mod);
    for (long long unsigned& x : a) {
      x = (long long unsigned)((1LL * x * n_1) % mod);
    }
  }
}
vector<long long unsigned> multiply(vector<long long unsigned> const& a,
                                    vector<long long unsigned> const& b) {
  vector<long long unsigned> fa(a.begin(), a.end()), fb(b.begin(), b.end());
  long long unsigned n = 1;
  while (n < a.size() + b.size()) n <<= 1;
  fa.resize(n);
  fb.resize(n);
  fft(fa, false);
  fft(fb, false);
  for (long long unsigned i = 0; i < n; i++) fa[i] *= fb[i];
  fft(fa, true);
  vector<long long unsigned> result(n);
  for (long long unsigned i = 0; i < n; i++) result[i] = fa[i];
  return result;
}
int main() {
  long long unsigned n;
  cin >> n;
  map<long long unsigned, long long unsigned> mp;
  map<long long unsigned, long long unsigned>::iterator it;
  for (long long unsigned i = 1; i <= n; i++) {
    long long unsigned x;
    cin >> x;
    mp[x]++;
  }
  it = mp.begin();
  vector<long long unsigned> a;
  a.push_back(1);
  for (long long unsigned i = 1; i <= it->second; i++) a.push_back(1);
  it++;
  while (it != mp.end()) {
    vector<long long unsigned> b;
    b.push_back(1);
    for (long long unsigned i = 1; i <= it->second; i++) b.push_back(1);
    a = multiply(a, b);
    it++;
  }
  long long unsigned mx = 0;
  cout << a[n / 2] % mod << endl;
  return 0;
}
