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;
long long gcd(long long x, long long y) {
  if (!x || !y) return x | y;
  if (x % y == 0)
    return y;
  else
    return gcd(y, x % y);
}
bool isprime(long long x) {
  if (x % 2 == 0) return 0;
  for (long long i = 3; i <= x / i; i = i + 2)
    if (x % i == 0) return 0;
  return 1;
}
bool check(set<int> meh, int m) {
  set<int>::iterator it;
  int count = 0;
  for (it = meh.begin(); it != meh.end(); it++)
    if (*it == m) return false;
  return true;
}
int main() {
  int t, r = 0;
  set<int> se, meh;
  cin >> t;
  for (int i = 0; i < t; i++) {
    int x = se.size();
    int m;
    cin >> m;
    se.insert(m);
    int y = se.size();
    if (x == y && check(meh, m)) r++, meh.insert(m);
  }
  int ans = (r * (r - 1) / 2) % 998244353;
  cout << (se.size() % 998244353 + max(0, ans)) % 998244353;
}
