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 INT = long long;
using pii = pair<int, int>;
using VI = vector<int>;
int power(int a, int b, int c, int ans = 1) {
  for (; b; b >>= 1, a = (INT)a * a % c)
    if (b & 1) ans = (INT)ans * a % c;
  return ans;
}
int a[201010], nn;
namespace fft {
const int mod = 998244353, root = 31, root_1 = power(root, mod - 2, mod);
void fft(VI &a, bool invert) {
  int n = a.size(), pw = power(invert ? root_1 : root, (mod - 1) / n, mod);
  for (int m = n, h; h = m / 2, m >= 2; pw = (INT)pw * pw % mod, m = h) {
    for (int i = 0, w = 1; i < h; i++, w = (INT)w * pw % mod) {
      for (int j = i; j < n; j += m) {
        int k = j + h, first = (a[j] - a[k] + mod) % mod;
        a[j] = (a[j] + a[k]) % mod;
        a[k] = (INT)w * first % mod;
      }
    }
  }
  for (int i = 0, j = 1; j < n - 1; j++) {
    for (int k = n >> 1; k > (i ^= k); k >>= 1)
      ;
    if (j < i) swap(a[i], a[j]);
  }
  if (invert) {
    int rev = power(n, mod - 2, mod);
    for (int i = 0; i < n; i++) a[i] = (INT)a[i] * rev % mod;
  }
}
VI multiply(const VI &a, const VI &b, int ok = 0) {
  int n = 1, mx = a.size() + b.size() - 1;
  while (n < mx) n <<= 1;
  if (mx < 128) {
    VI c(mx);
    for (int i = 0; i < a.size(); i++)
      for (int j = 0; j < b.size(); j++)
        c[i + j] = (c[i + j] + (INT)a[i] * b[j]) % mod;
    if (c.size() > nn + 1) c.resize(nn + 1);
    return c;
  }
  VI na = a, nb = b;
  na.resize(n);
  nb.resize(n);
  fft(na, false);
  fft(nb, false);
  for (int i = 0; i < n; i++) na[i] = (INT)na[i] * nb[i] % mod;
  fft(na, true);
  na.resize(mx);
  if (na.size() > nn + 1) na.resize(nn + 1);
  return na;
}
};  // namespace fft
struct cmp {
  bool operator()(const VI &a, const VI &b) { return a.size() > b.size(); }
};
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  sort(a, a + n);
  int m = 0;
  nn = n / 2;
  priority_queue<VI, vector<VI>, cmp> pq;
  for (int i = 0, j; i < n; i = j) {
    for (j = i + 1; j < n && a[j] == a[i]; j++)
      ;
    pq.push(VI(min(nn, j - i) + 1, 1));
  }
  while (pq.size() > 1) {
    VI a = pq.top();
    pq.pop();
    VI b = pq.top();
    pq.pop();
    pq.push(fft::multiply(a, b));
  }
  cout << pq.top()[n / 2] << endl;
  return 0;
}
