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,no-stack-protector,unroll-all-loops,unroll-loops")
#pragma GCC target("abm,mmx,sse,sse2,sse3,ssse3,sse4a,popcnt,tune=native")
using namespace std;
const int INF = 1e9;
const long long MOD = 998244353;
long long f[1000000];
long long my_pow(long long a, long long b) {
  if (b == 0) return 1;
  long long p = my_pow(a, b / 2);
  if (b % 2 == 0)
    return p * p % MOD;
  else
    return (p * p % MOD) * a % MOD;
}
int C(int k, int n) {
  return f[n] * (my_pow(f[k], MOD - 2) * my_pow(f[n - k], MOD - 2) % MOD) % MOD;
}
map<int, int> cnt;
int32_t main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int p;
    cin >> p;
    cnt[p]++;
  }
  int res = INF;
  for (pair<int, int> a : cnt) res = min(res, a.second);
  int k = (int)cnt.size();
  f[0] = 1;
  for (int i = 1; i <= n; i++) f[i] = ((long long)i * f[i - 1]) % MOD;
  cout << C(res, res + k - 1) << endl;
  return 0;
}
