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 = 2e5 + 5;
const int maxm = 3e6 + 5;
int s[maxn], sum[maxn], cnt[maxm];
vector<int> A, B;
const long long mod = 998244353, modg = 3, modi = 332748118;
int a[maxn], b[maxn];
int pow_mod(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = (long long)ans * a % mod;
    a = (long long)a * a % mod, b >>= 1;
  }
  return ans;
}
int add(int a, int b) {
  a += b;
  if (a > mod) return a - mod;
  return a;
}
int sub(int a, int b) {
  a -= b;
  if (a < 0) return a + mod;
  return a;
}
const int maxl = 2e5 + 10;
class nubmer {
 public:
  int rev[maxl << 2], len, pw;
  long long wt[2][maxl << 2];
  void init_0() {
    int len = 1;
    while (len <= maxl) len <<= 1;
    for (int mid = 1; mid < len; mid <<= 1) {
      long long wn1 = pow_mod(modg, (mod - 1) / (mid << 1));
      long long wn2 = pow_mod(modi, (mod - 1) / (mid << 1));
      wt[0][mid] = wt[1][mid] = 1;
      long long wt1 = wn1, wt2 = wn2;
      for (int j = 1; j <= mid; j++) {
        wt[0][mid + j] = wt1;
        wt1 = wt1 * wn1 % mod;
        wt[1][mid + j] = wt2;
        wt2 = wt2 * wn2 % mod;
      }
    }
  }
  void transform(vector<int> &a, int flag) {
    long long *f = flag == 1 ? wt[0] : wt[1];
    if (a.size() != len) a.resize(len);
    for (int i = 0; i <= len - 1; i++)
      if (i < rev[i]) swap(a[i], a[rev[i]]);
    for (int mid = 1; mid < len; mid <<= 1) {
      for (int r = mid << 1, j = 0; j < len; j += r) {
        long long *p = f + mid;
        for (int k = 0; k < mid; ++k, ++p) {
          int x = a[j + k], y = (*p) * a[j + k + mid] % mod;
          a[j + k + mid] = sub(a[j + k], y), a[j + k] += y;
          if (a[j + k] > mod) a[j + k] -= mod;
        }
      }
    }
    if (flag == -1) {
      long long inv = pow_mod(len, mod - 2);
      for (int i = 0; i <= len - 1; i++) {
        a[i] = a[i] * inv % mod;
        if (a[i] < 0) a[i] += mod;
      }
    }
  }
  void fix(vector<int> &a) {
    while (!a.empty() && !a.back()) a.pop_back();
  }
  vector<int> mul(vector<int> &a, vector<int> &b) {
    int la = a.size(), lb = b.size();
    if (la + lb <= 100) {
      vector<int> c(la + lb, 0);
      for (int i = 0; i <= la - 1; i++)
        for (int j = 0; j <= lb - 1; j++)
          c[i + j] = (c[i + j] + (long long)a[i] * b[j]) % mod;
      return fix(c), c;
    }
    int n = la + lb;
    len = 1, pw = 0;
    while (len <= n) len <<= 1, pw++;
    for (int i = 0; i <= len - 1; i++)
      rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (pw - 1));
    transform(a, 1);
    transform(b, 1);
    vector<int> c(len, 0);
    for (int i = 0; i <= len - 1; i++) c[i] = (long long)a[i] * b[i] % mod;
    transform(c, -1);
    return fix(c), c;
  }
} ntt;
vector<int> solve(int l, int r) {
  if (l == r) {
    vector<int> ans(s[l] + 1, 1);
    return ans;
  }
  int mid = (l + r) >> 1;
  vector<int> a = solve(l, mid), b = solve(mid + 1, r);
  int sz = a.size() + b.size();
  A.resize(sz + 1);
  B.resize(sz + 1);
  for (int i = 0; i < sz; i++) A[i] = B[i] = 0;
  for (int i = 0; i < a.size(); i++) A[i] = a[i];
  for (int i = 0; i < b.size(); i++) B[i] = b[i];
  vector<int> now = ntt.mul(A, B);
  int len = sum[r] - sum[l - 1];
  vector<int> ans(len + 1);
  for (int i = 0; i <= len; i++) ans[i] = now[i];
  return ans;
}
int main() {
  int n, len = 0;
  cin >> n;
  ntt.init_0();
  for (int i = 1; i <= n; i++) {
    int t;
    cin >> t;
    cnt[t]++;
  }
  for (int i = 2; i < maxm; i++) {
    if (cnt[i]) {
      s[++len] = cnt[i];
      sum[len] = sum[len - 1] + cnt[i];
    }
  }
  vector<int> ans = solve(1, len);
  cout << ans[n / 2] << '\n';
}
