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>
int a[100001];
int tree[4 * 100001];
int n;
void update(int node, int b, int e, int idx, int value) {
  if (idx < b || idx > e) return;
  if (idx <= b && idx >= e) {
    tree[node] += value;
    return;
  }
  int mid = (b + e) / 2;
  int left = 2 * node, right = 2 * node + 1;
  update(left, b, mid, idx, value);
  update(right, mid + 1, e, idx, value);
  tree[node] = tree[left] + tree[right];
}
int query(int node, int b, int e, int k) {
  if (b == e) {
    return b;
  }
  int mid = (b + e) / 2;
  int left = 2 * node, right = 2 * node + 1;
  int store = tree[left];
  int ans = 0, lft = 0, rgt = 0;
  if (store < k) return query(right, mid + 1, e, k - store);
  return query(left, b, mid, k);
}
int main() {
  int T;
  scanf("%d", &T);
  int n = 0, ans = 0;
  int test_case;
  for (test_case = 1; test_case <= T; ++test_case) {
    int t, k;
    scanf("%d %d", &t, &k);
    if (t == 1) {
      n++;
      update(1, 1, 100001 - 1, k, 1);
    } else if (t == 2) {
      n--;
      update(1, 1, 100001 - 1, k, -1);
    } else if (t == 3) {
      ans = query(1, 1, 100001 - 1, k);
      printf("%d\n", ans);
    } else if (t == 4) {
      int pathaw = (n - k + 1);
      ans = query(1, 1, 100001 - 1, pathaw);
      printf("%d\n", ans);
    }
  }
  return 0;
}
