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;
int p[3000100], cnt[3000100], arr[3000100];
vector<int> ans, A;
namespace fft {
const int mod = 998244353;
const int root = 31;
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
const int root_1 = power(root, mod - 2, mod);
using INT = long long;
void fft(vector<int>& a, bool invert) {
  int n = a.size();
  int PW = power(invert ? root_1 : root, (mod - 1) / n, mod);
  for (int m = n, h; h = m / 2, m >= 2; PW = 1LL * PW * PW % mod, m = h) {
    for (int i = 0, w = 1; i < h; ++i, w = 1LL * w * PW % mod)
      for (int j = i; j < n; j += m) {
        int k = j + h, x = (a[j] - a[k] + mod) % mod;
        a[j] += a[k];
        a[j] %= mod;
        a[k] = 1LL * w * x % mod;
      }
  }
  for (int i = 0, j = 1; j < n - 1; ++j) {
    for (int k = n / 2; k > (i ^= k); k /= 2)
      ;
    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] = 1LL * a[i] * rev % mod;
  }
}
vector<int> multiply(const vector<int>& a, const vector<int>& b, int ok = 0) {
  int n = 1, mx = a.size() + b.size() - 1;
  while (n < mx) n <<= 1;
  if (mx < 256) {
    vector<int> c(mx);
    for (int i = 0; i < a.size(); i++)
      for (int j = 0; j < b.size(); j++) {
        c[i + j] = (c[i + j] + 1ll * a[i] * b[j]) % mod;
      }
    return c;
  }
  vector<int> na = a, nb = b;
  na.resize(n);
  nb.resize(n);
  fft(na, false);
  if (ok)
    nb = na;
  else
    fft(nb, false);
  for (int i = 0; i < n; ++i) na[i] = 1LL * na[i] * nb[i] % mod;
  fft(na, true);
  na.resize(mx);
  return na;
}
vector<int> findInverse(const vector<int>& ff) {
  int nn = ff.size();
  int n = 1;
  while (n < nn) n <<= 1;
  vector<int> f = ff;
  f.resize(n);
  vector<int> r(1, power(f[0], mod - 2, mod));
  for (int it = 2; it <= n; it <<= 1) {
    vector<int> nr =
        multiply(multiply(r, r), vector<int>(f.begin(), f.begin() + it));
    nr.resize(it);
    for (int i = 0; i < nr.size(); ++i)
      if (nr[i]) nr[i] = mod - nr[i];
    for (int i = 0; i < it / 2; ++i) nr[i] = (2LL * r[i] + nr[i]) % mod;
    r = nr;
  }
  r.resize(nn);
  return r;
}
bool checkInverse(const vector<int>& a, const vector<int>& b) {
  for (int i = 0; i < a.size(); i++) {
    int val = 0;
    for (int j = 0; j <= i; j++) {
      val = (val + 1LL * a[i - j] * b[j]) % mod;
    }
    if (val != (i == 0)) return false;
  }
  return true;
}
vector<int> squareRoot(const vector<int>& ff) {
  int nn = ff.size();
  int n = 1;
  while (n < nn) n <<= 1;
  vector<int> f = ff;
  f.resize(n);
  vector<int> r(1, 1);
  int inv2 = power(2, mod - 2, mod);
  for (int it = 2; it <= n; it <<= 1) {
    r.resize(it);
    vector<int> res =
        multiply(findInverse(r), vector<int>(f.begin(), f.begin() + it));
    res.resize(it);
    for (int i = 0; i < r.size(); ++i) {
      res[i] += r[i];
      if (res[i] >= mod) res[i] -= mod;
    }
    for (int i = 0; i < res.size(); ++i) res[i] = 1LL * res[i] * inv2 % mod;
    r = res;
  }
  r.resize(nn);
  return r;
}
};  // namespace fft
vector<int> calc(vector<int> vec) {
  vector<int> tmp;
  if (vec.size() == 1) {
    for (int i = 0; i <= vec[0]; i++) tmp.push_back(1);
    return tmp;
  }
  int s = 0;
  for (int i = 0; i < vec.size(); i++) s += vec[i];
  vector<int> a, b;
  int c = vec[0];
  a.push_back(vec[0]);
  for (int i = 1; i < vec.size(); i++) {
    c += vec[i];
    if (2 * c <= s)
      a.push_back(vec[i]);
    else
      b.push_back(vec[i]);
  }
  return fft::multiply(calc(a), calc(b));
}
int main() {
  int n, nn = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
    cnt[p[i]]++;
    arr[++nn] = p[i];
  }
  sort(arr + 1, arr + nn + 1);
  nn = unique(arr + 1, arr + nn + 1) - arr - 1;
  vector<int> vec;
  for (int i = 1; i <= nn; i++) vec.push_back(cnt[arr[i]]);
  ans = calc(vec);
  cout << ans[n / 2] << endl;
  return 0;
}
