You are given a sequence of n pairs of integers: (a_1, b_1), (a_2, b_2), ... , (a_n, b_n). This sequence is called bad if it is sorted in non-descending order by first elements or if it is sorted in non-descending order by second elements. Otherwise the sequence is good. There are examples of good and bad sequences:

  * s = [(1, 2), (3, 2), (3, 1)] is bad because the sequence of first elements is sorted: [1, 3, 3]; 
  * s = [(1, 2), (3, 2), (1, 2)] is bad because the sequence of second elements is sorted: [2, 2, 2]; 
  * s = [(1, 1), (2, 2), (3, 3)] is bad because both sequences (the sequence of first elements and the sequence of second elements) are sorted; 
  * s = [(1, 3), (3, 3), (2, 2)] is good because neither the sequence of first elements ([1, 3, 2]) nor the sequence of second elements ([3, 3, 2]) is sorted. 



Calculate the number of permutations of size n such that after applying this permutation to the sequence s it turns into a good sequence. 

A permutation p of size n is a sequence p_1, p_2, ... , p_n consisting of n distinct integers from 1 to n (1 ≤ p_i ≤ n). If you apply permutation p_1, p_2, ... , p_n to the sequence s_1, s_2, ... , s_n you get the sequence s_{p_1}, s_{p_2}, ... , s_{p_n}. For example, if s = [(1, 2), (1, 3), (2, 3)] and p = [2, 3, 1] then s turns into [(1, 3), (2, 3), (1, 2)].

Input

The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5).

The next n lines contains description of sequence s. The i-th line contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ n) — the first and second elements of i-th pair in the sequence.

The sequence s may contain equal elements.

Output

Print the number of permutations of size n such that after applying this permutation to the sequence s it turns into a good sequence. Print the answer modulo 998244353 (a prime number).

Examples

Input


3
1 1
2 2
3 1


Output


3


Input


4
2 3
2 2
2 1
2 4


Output


0


Input


3
1 1
1 1
2 3


Output


4

Note

In first test case there are six permutations of size 3: 

  1. if p = [1, 2, 3], then s = [(1, 1), (2, 2), (3, 1)] — bad sequence (sorted by first elements); 
  2. if p = [1, 3, 2], then s = [(1, 1), (3, 1), (2, 2)] — bad sequence (sorted by second elements); 
  3. if p = [2, 1, 3], then s = [(2, 2), (1, 1), (3, 1)] — good sequence; 
  4. if p = [2, 3, 1], then s = [(2, 2), (3, 1), (1, 1)] — good sequence; 
  5. if p = [3, 1, 2], then s = [(3, 1), (1, 1), (2, 2)] — bad sequence (sorted by second elements); 
  6. if p = [3, 2, 1], then s = [(3, 1), (2, 2), (1, 1)] — good sequence. 
#include <bits/stdc++.h>
using namespace std;
const int N = 300010;
class cmp {
 public:
  bool operator()(const pair<long long int, long long int> &a,
                  const pair<long long int, long long int> &b) {
    if (a.first == b.first) return a.second > b.second;
    return a.first < b.first;
  }
};
long long int a[N], b[N], fact[N];
long long int mod = 998244353;
void cal() {
  fact[1] = 1;
  for (long long int i = 2; i < N; i++) {
    fact[i] = ((fact[i - 1] % mod) * (i % mod)) % mod;
  }
}
bool sort1(const pair<long long int, long long int> &x,
           const pair<long long int, long long int> &y) {
  if (x.first == y.first) return x.second < y.second;
  return x.first < y.first;
}
void solve() {
  long long int n;
  cin >> n;
  cal();
  memset(a, 0, sizeof(a));
  memset(b, 0, sizeof(b));
  long long int i, x, y, f = 0;
  vector<pair<long long int, long long int>> v;
  for (i = 0; i < n; i++) {
    cin >> x >> y;
    a[x]++;
    b[y]++;
    v.push_back({x, y});
    if (a[x] == n || b[y] == n) f = 1;
  }
  if (f) {
    cout << "0" << endl;
    return;
  }
  long long int ans = fact[n];
  long long int ans1 = 1;
  for (i = 1; i < N; i++) {
    if (a[i]) {
      long long int cnt = fact[a[i]];
      ans1 = ((ans1 % mod) * (cnt % mod)) % mod;
    }
  }
  ans = (ans - ans1 + mod) % mod;
  long long int ans2 = 1;
  for (i = 1; i < N; i++) {
    if (b[i]) {
      long long int cnt = fact[b[i]];
      ans2 = ((ans2 % mod) * (cnt % mod)) % mod;
    }
  }
  ans = (ans - ans2 + mod) % mod;
  sort(v.begin(), v.end(), sort1);
  long long int k = 0;
  for (i = 0; i < n - 1; i++) {
    if (v[i].second > v[i + 1].second) {
      k = 1;
      break;
    }
  }
  if (k == 0) {
    ans1 = 1;
    for (i = 0; i < n;) {
      long long int j = i;
      long long int cnt = 0;
      while (j < n && v[i].first == v[j].first && v[i].second == v[j].second) {
        j++;
        cnt++;
      }
      i = j;
      cnt = fact[cnt];
      ans1 = ((ans1 % mod) * (cnt % mod)) % mod;
    }
    ans = (ans + ans1) % mod;
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
