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;
map<int, int> mp;
int n;
namespace fft {
const int mod = 998244353;
int mPow(int a, int x, int ans = 1) {
  for (; x; x >>= 1, a = (long long)a * a % mod)
    if (x & 1) ans = (long long)ans * a % mod;
  return ans;
}
const int root = 31;
const int root_1 = mPow(root, mod - 2);
void fft(vector<int> &a, bool invert) {
  int n = a.size();
  int PW = mPow(invert ? root_1 : root, (mod - 1) / n);
  for (int m = n, h; h = m / 2, m >= 2; PW = (long long)PW * PW % mod, m = h) {
    for (int i = 0, w = 1; i < h; i++, w = (long long)w * PW % mod)
      for (int j = i, k, x; j < n; j += m) {
        k = j + h;
        ((x = a[j] - a[k]) < 0) && (x += mod);
        ((a[j] = a[j] + a[k]) >= mod) && (a[j] -= mod);
        a[k] = (long long)w * x % mod;
      }
  }
  for (int i = 0, j = 1; j < n - 1; j++) {
    for (int k = n / 2; k > (i ^= k); k >>= 1)
      ;
    if (j < i) swap(a[i], a[j]);
  }
  if (invert) {
    int rev = mPow(n, mod - 2);
    for (int i = 0; i < n; i++) {
      a[i] = (long long)a[i] * rev % mod;
    }
  }
}
vector<int> multiply(const vector<int> &a, const vector<int> &b) {
  int n = 1 << (32 - __builtin_clz(a.size() + b.size() - 1));
  vector<int> 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] = (long long)na[i] * nb[i] % mod;
  fft(na, true);
  return na;
}
};  // namespace fft
vector<int> solve(vector<int> &buf) {
  if (buf.size() == 1) {
    vector<int> A = vector<int>(buf[0] + 1, 1);
    return A;
  }
  int sum = 0;
  for (auto p : buf) sum += p;
  int tot = 0;
  vector<int> A, B;
  for (int i = 0; i < buf.size(); i++) {
    tot += buf[i];
    if (tot + tot > sum)
      B.push_back(buf[i]);
    else
      A.push_back(buf[i]);
  }
  return fft::multiply(solve(A), solve(B));
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int u;
    scanf("%d", &u);
    mp[u]++;
  }
  vector<int> buf;
  for (auto p : mp) {
    buf.push_back(p.second);
  }
  sort(buf.begin(), buf.end());
  auto p = solve(buf);
  cout << p[n / 2] << endl;
}
