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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <typename T1, typename T2>
inline void chkmin(T1 &first, T2 second) {
  if (first > second) first = second;
}
template <typename T1, typename T2>
inline void chkmax(T1 &first, T2 second) {
  if (first < second) first = second;
}
template <typename T, typename U>
inline ostream &operator<<(ostream &_out, const pair<T, U> &_p) {
  _out << _p.first << ' ' << _p.second;
  return _out;
}
template <typename T, typename U>
inline istream &operator>>(istream &_in, pair<T, U> &_p) {
  _in >> _p.first >> _p.second;
  return _in;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const vector<T> &_v) {
  if (_v.empty()) {
    return _out;
  }
  _out << _v.front();
  for (auto _it = ++_v.begin(); _it != _v.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T>
inline istream &operator>>(istream &_in, vector<T> &_v) {
  for (auto &_i : _v) {
    _in >> _i;
  }
  return _in;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const set<T> &_s) {
  if (_s.empty()) {
    return _out;
  }
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const multiset<T> &_s) {
  if (_s.empty()) {
    return _out;
  }
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const unordered_set<T> &_s) {
  if (_s.empty()) {
    return _out;
  }
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const unordered_multiset<T> &_s) {
  if (_s.empty()) {
    return _out;
  }
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T, typename U>
inline ostream &operator<<(ostream &_out, const map<T, U> &_m) {
  if (_m.empty()) {
    return _out;
  }
  _out << '(' << _m.begin()->first << ": " << _m.begin()->second << ')';
  for (auto _it = ++_m.begin(); _it != _m.end(); ++_it) {
    _out << ", (" << _it->first << ": " << _it->second << ')';
  }
  return _out;
}
template <typename T, typename U>
inline ostream &operator<<(ostream &_out, const unordered_map<T, U> &_m) {
  if (_m.empty()) {
    return _out;
  }
  _out << '(' << _m.begin()->first << ": " << _m.begin()->second << ')';
  for (auto _it = ++_m.begin(); _it != _m.end(); ++_it) {
    _out << ", (" << _it->first << ": " << _it->second << ')';
  }
  return _out;
}
const int mod = 998244353;
const int root = 31;
int mul(long long a, int b) { return a * b % mod; }
int powx(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = mul(ans, a);
    a = mul(a, a);
    b >>= 1;
  }
  return ans;
}
const int root_1 = powx(root, mod - 2);
const int root_pw = 1 << 23;
void fft(vector<int> &a, bool invert) {
  int n = (int)a.size();
  for (int i = 1, j = 0; i < n; ++i) {
    int bit = n >> 1;
    for (; j >= bit; bit >>= 1) j -= bit;
    j += bit;
    if (i < j) swap(a[i], a[j]);
  }
  for (int len = 2; len <= n; len <<= 1) {
    int wlen = invert ? root_1 : root;
    for (int i = len; i < root_pw; i <<= 1) wlen = int(wlen * 1ll * wlen % mod);
    for (int i = 0; i < n; i += len) {
      int w = 1;
      for (int j = 0; j < len / 2; ++j) {
        int u = a[i + j], v = int(a[i + j + len / 2] * 1ll * w % mod);
        a[i + j] = u + v < mod ? u + v : u + v - mod;
        a[i + j + len / 2] = u - v >= 0 ? u - v : u - v + mod;
        w = int(w * 1ll * wlen % mod);
      }
    }
  }
  if (invert) {
    int nrev = powx(n, mod - 2);
    for (int i = 0; i < n; ++i) a[i] = int(a[i] * 1ll * nrev % mod);
  }
}
vector<int> multiply(const vector<int> a, const vector<int> b) {
  vector<int> fa(a.begin(), a.end()), fb(b.begin(), b.end());
  size_t n = 1;
  while (n < max(a.size(), b.size())) n <<= 1;
  n <<= 1;
  fa.resize(n), fb.resize(n);
  fft(fa, false), fft(fb, false);
  for (size_t i = 0; i < n; ++i) fa[i] = 1LL * fa[i] * fb[i] % mod;
  fft(fa, true);
  return fa;
}
const int N = 4e6;
int n;
vector<int> cnt;
int a[N];
vector<int> solve(int L, int R) {
  if (R - L == 1) {
    return vector<int>(cnt[L] + 1, 1);
  }
  int M = L;
  int sum = 0;
  for (int i = L; i < R; ++i) {
    sum += cnt[i];
  }
  int cur = 0;
  while ((cur + cnt[M]) * 2 <= sum) {
    cur += cnt[M];
    ++M;
  }
  return multiply(solve(L, M), solve(M, R));
}
int main() {
  ios::sync_with_stdio(0);
  srand(time(0));
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int p;
    cin >> p;
    ++a[p];
  }
  for (int i = 0; i < N; ++i) {
    if (a[i]) {
      cnt.push_back(a[i]);
    }
  }
  sort(cnt.begin(), cnt.end());
  auto res = solve(0, cnt.size());
  cout << res[n / 2] << '\n';
}
