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 gen(chrono::system_clock::now().time_since_epoch().count());
const int N = 2e5 + 7;
const int M = 1e9 + 7;
const int FFTM = 998244353;
int n, x;
map<int, int> all;
vector<int> p;
int lp[N], f[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 2; i <= n; i++) {
    if (!lp[i]) {
      lp[i] = i;
      p.push_back(i);
    }
    for (int j = 0; j < p.size() && lp[i] >= p[j] && i * p[j] < N; j++)
      lp[i * p[j]] = p[j];
  }
  f[1] = 1;
  f[2] = 1;
  for (int i = 3; i <= n; i++) f[i] = f[i - 1] + (lp[i] == i);
  for (int i = 0; i < n; i++) cin >> x, all[x]++;
  long long res = 0;
  long long add = 0;
  int was = 0;
  int fr = 0;
  for (auto it : all) {
    if (f[it.second] != 1) was = 1;
    if (was) {
      if (fr)
        res *= f[it.second];
      else
        res = f[it.second];
      fr = 1;
    }
    add += (f[it.second] == 1);
    res %= FFTM;
  }
  cout << (res + add) % FFTM;
  return 0;
}
