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;
template <class T1, class T2>
istream &operator>>(istream &in, pair<T1, T2> &P) {
  in >> P.first >> P.second;
  return in;
}
template <class T1, class T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &P) {
  out << "(" << P.first << ", " << P.second << ")";
  return out;
}
template <class T>
istream &operator>>(istream &in, vector<T> &arr) {
  for (auto &x : arr) in >> x;
  return in;
}
template <class T>
ostream &operator<<(ostream &out, const vector<T> &arr) {
  for (auto &x : arr) out << x << ' ';
  cout << "\n";
  return out;
}
template <class T>
istream &operator>>(istream &in, deque<T> &arr) {
  for (auto &x : arr) in >> x;
  return in;
}
template <class T>
ostream &operator<<(ostream &out, const deque<T> &arr) {
  for (auto &x : arr) out << x << ' ';
  cout << "\n";
  return out;
}
int n;
map<int, int> pr;
vector<vector<long long> > poly;
const long long mod = (119 << 23) + 1, root = 62;
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;
}
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};
}
vector<long long> solve(int l, int r) {
  int mid = (l + r) / 2;
  if (l == r) {
    return poly[l];
  }
  vector<long long> a = solve(l, mid);
  vector<long long> b = solve(mid + 1, r);
  return conv(a, b);
}
int32_t main() {
  scanf("%d", &n);
  for (int i = 0; i < (n); ++i) {
    int x;
    scanf("%d", &x);
    pr[x]++;
  }
  for (auto w : pr) {
    vector<long long> curr;
    curr.assign(w.second + 1, 1);
    poly.push_back(curr);
  }
  vector<long long> ans = solve(0, ((int)poly.size()) - 1);
  printf("%lld\n", ans[n / 2]);
}
