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;
int a[200020];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  sort(a, a + n);
  int m = 0;
  for (int i = 0, j; i < n; i = j) {
    for (j = i + 1; j < n && a[j] == a[i]; j++)
      ;
    a[m++] = j - i;
  }
  sort(a, a + m);
  if (m == 1) {
    puts("1");
    return 0;
  }
  if (m == 2) {
    printf("%d\n", a[0] + 1);
    return 0;
  }
  long long ans = 0, mod = 998244353;
  for (int i = 0; i < m - 1; i++) {
    ans += (long long)a[i] * (m - 1 - i) % mod;
  }
  cout << ans % mod << endl;
}
