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;
using cd = complex<double>;
const double PI = acos(-1);
void fft(vector<cd>& a, bool invert) {
  long long unsigned n = a.size();
  if (n == 1) return;
  vector<cd> a0(n / 2), a1(n / 2);
  for (long long unsigned i = 0; 2 * i < n; i++) {
    a0[i] = a[2 * i];
    a1[i] = a[2 * i + 1];
  }
  fft(a0, invert);
  fft(a1, invert);
  double ang = 2 * PI / n * (invert ? -1 : 1);
  cd w(1), wn(cos(ang), sin(ang));
  for (long long unsigned i = 0; 2 * i < n; i++) {
    a[i] = a0[i] + w * a1[i];
    a[i + n / 2] = a0[i] - w * a1[i];
    if (invert) {
      a[i] /= 2;
      a[i + n / 2] /= 2;
    }
    w *= wn;
  }
}
vector<long long unsigned> multiply(vector<long long unsigned> const& a,
                                    vector<long long unsigned> const& b) {
  vector<cd> 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] = round(fa[i].real());
  return result;
}
const long long unsigned mod = 998244353;
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;
}
