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;
const int mod = 998244353;
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;
}
int r[200010], p[200010];
namespace fft {
const int mod = 998244353;
const int root = 31;
const int root_1 = power(root, mod - 2, mod);
int mPow(int a, int x) { return power(a, x, mod); }
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 = 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 = mPow(n, mod - 2);
    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 n = max(a.size(), b.size());
  vector<int> na = a, nb = b;
  na.resize(2 * n);
  nb.resize(2 * n);
  fft(na, false);
  fft(nb, false);
  for (int i = 0; i < 2 * n; ++i) na[i] = 1LL * na[i] * nb[i] % mod;
  fft(na, true);
  return na;
}
vector<int> findInverse(const vector<int>& f) {
  int n = f.size();
  vector<int> r(1, mPow(f[0], mod - 2));
  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) nr[i] = mod - nr[i];
    for (int i = 0; i < it / 2; ++i) nr[i] = (2LL * r[i] + nr[i]) % mod;
    r = nr;
  }
  return r;
}
};  // namespace fft
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
  }
  sort(p + 1, p + n + 1);
  int k = 0;
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    int st = i;
    while (p[st] == p[i] && st <= n) st++;
    st--;
    r[++k] = st - i + 2;
    sum += r[k] - 1;
    i = st;
  }
  sort(r + 1, r + k + 1);
  vector<int> a, f;
  a.resize(4);
  a[0] = mod - 1;
  a[1] = 1;
  f.resize(4);
  f[0] = 1;
  int cnt = k;
  while (cnt) {
    if (cnt & 1) f = fft ::multiply(f, a);
    a = fft ::multiply(a, a);
    cnt >>= 1;
  }
  f = fft ::findInverse(f);
  for (int i = 1; i <= k; i++) {
    vector<int> A;
    int len = 1;
    while (len <= r[i]) len *= 2;
    A.resize(len);
    A[r[i]] = 1;
    A[0] = -1;
    f = fft ::multiply(f, A);
  }
  printf("%d\n", f[n / 2]);
}
