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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long mod = 998244353;
int LIMIT;
long long modpow(long long b, long long e) {
  long long ans = 1;
  for (; e; b = b * b % mod, e /= 2)
    if (e & 1) ans = ans * b % mod;
  return ans;
}
const long long root = 62;
inline void ntt(vector<long long>& a, vector<long long>& rt,
                vector<long long>& rev, int n) {
  for (int i = 0; i < (n); ++i)
    if (i < rev[i]) swap(a[i], a[rev[i]]);
  for (int k = 1; k < n; k *= 2)
    for (int i = 0; i < n; i += 2 * k)
      for (int j = 0; j < (k); ++j) {
        long long z = rt[j + k] * a[i + j + k] % mod, &ai = a[i + j];
        a[i + j + k] = (z > ai ? ai - z + mod : ai - z);
        ai += (ai + z >= mod ? z - mod : z);
      }
}
vector<long long> conv(const vector<long long>& a, const vector<long long>& b) {
  if (a.empty() || b.empty()) return {};
  int s = (int)(a).size() + (int)(b).size() - 1, B = 32 - __builtin_clz(s),
      n = 1 << B;
  vector<long long> L(a), R(b), out(n), rt(n, 1), rev(n);
  L.resize(n), R.resize(n);
  for (int i = 0; i < (n); ++i) rev[i] = (rev[i / 2] | (i & 1) << B) / 2;
  long long curL = mod / 2, inv = modpow(n, mod - 2);
  for (int k = 2; k < n; k *= 2) {
    long long z[] = {1, modpow(root, curL /= 2)};
    for (int i = k; i < (2 * k); ++i) rt[i] = rt[i / 2] * z[i & 1] % mod;
  }
  ntt(L, rt, rev, n);
  ntt(R, rt, rev, n);
  for (int i = 0; i < (n); ++i)
    out[-i & (n - 1)] = L[i] * R[i] % mod * inv % mod;
  ntt(out, rt, rev, n);
  return {out.begin(), out.begin() + s};
}
void read(int& n) {
  n = 0;
  char c = getchar();
  while (c >= '0' && c <= '9') {
    n = n * 10 + c - '0';
    c = getchar();
  }
}
int cnt[3000005];
vector<long long> t;
vector<long long> res;
vector<int> s;
int a[200005];
int main() {
  int n;
  read(n);
  LIMIT = n / 2 + 1;
  for (int i = 1; i <= n; i++) {
    read(a[i]);
    cnt[a[i]]++;
  }
  for (int i = 1; i <= n; i++) {
    if (cnt[a[i]]) {
      s.push_back(cnt[a[i]] + 1);
      cnt[a[i]] = 0;
    }
  }
  sort(s.begin(), s.end());
  int sz = 0, s2;
  for (auto& x : s) {
    while (sz < x) {
      t.push_back(1);
      ++sz;
    }
    if (res.empty()) {
      res = t;
    } else {
      if (res.size() > t.size()) swap(res, t);
      res = conv(res, t);
    }
    s2 = res.size();
    while (s2 > LIMIT) {
      res.pop_back();
      s2--;
    }
  }
  cout << res[n / 2];
}
