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::literals::string_literals;
using i64 = std::int_fast64_t;
using std::cout;
using std::cerr;
using std::endl;
using std::cin;

template<typename T>
std::vector<T> make_v(size_t a){return std::vector<T>(a);}

template<typename T,typename... Ts>
auto make_v(size_t a,Ts... ts){
  return std::vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));
}

#define NDEBUG

int main() {
	int n; scanf("%d", &n); std::vector<int> a(n);
	for(int i = 0; i < n; i++) scanf("%d", &a[i]);

	// solve
	std::map<int, int> mp;
	for(auto v: a) mp[v]++;

	i64 A = 0, B = 0;
	for(auto p: mp) {
		A += (p.second == 1);
		B += (p.second != 1);
	}

	#ifndef NDEBUG
	cerr << A << " " << B << endl;
	#endif

	i64 ans = (A + B) * (A + B - 1) / 2LL + B;
	printf("%lld\n", ans);
	return 0;
}
