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;
const int MAXN = 200200;
const int MOD = 998244353;
int add(int x, int y) {
  x += y;
  if (x >= MOD) {
    x -= MOD;
  }
  return x;
}
int mul(int x, int y) { return (long long)x * y % MOD; }
int pw(int x, int y) {
  int ans = 1;
  for (; y > 0; y >>= 1, x = mul(x, x)) {
    if (y & 1) {
      ans = mul(ans, x);
    }
  }
  return ans;
}
int cnt[MAXN];
int fact[MAXN];
int ifact[MAXN];
int C(int n, int k) {
  if (n < k || k < 0) {
    return 0;
  } else {
    return mul(fact[n], mul(ifact[k], ifact[n - k]));
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int N;
  cin >> N;
  map<int, int> facs;
  for (int i = 0; i < N; ++i) {
    int x;
    cin >> x;
    ++facs[x];
  }
  for (auto z : facs) {
    ++cnt[z.second];
  }
  fact[0] = 1;
  for (int i = 1; i < MAXN; ++i) {
    fact[i] = mul(fact[i - 1], i);
  }
  ifact[MAXN - 1] = pw(fact[MAXN - 1], MOD - 2);
  for (int i = MAXN - 1; i >= 0; --i) {
    ifact[i - 1] = mul(ifact[i], i);
  }
  int alive = facs.size();
  int ans = 0;
  for (int z = 1; z <= alive; ++z) {
    ans = add(ans, C(alive, z));
    alive -= cnt[z];
  }
  cout << ans << "\n";
  return 0;
}
