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 long long M = 1e9 + 7;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  long long m = 0;
  vector<long long> lol(26, 0);
  vector<vector<long long>> kek(1e6 + 1, lol);
  long long ans = 0;
  for (long long _ = 0; _ < n; _++) {
    string s;
    cin >> s;
    for (long long i = 0; i < s.size(); i++) {
      kek[i][s[i] - 'a']++;
    }
    ans += s.size();
  }
  for (long long j = 0; j <= 1e6; j++) {
    long long mx = 0;
    for (long long x = 0; x < 26; x++) {
      mx = max(mx, kek[j][x]);
    }
    ans -= mx;
  }
  cout << ans;
}
