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 MAXN = 1000 * 100 * 3 + 10;
long long int mod = 998244353;
long long int x[MAXN];
long long int y[MAXN];
pair<long long int, long long int> p[MAXN];
long long int f(long long int n) {
  long long int m = 1;
  for (int i = 1; i <= n; i++) {
    m *= i % mod;
    m = m % mod;
  }
  return m % mod;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x[i] >> y[i];
    p[i].first = x[i];
    p[i].second = y[i];
  }
  sort(x, x + n);
  sort(y, y + n);
  sort(p, p + n);
  long long int k1 = 1;
  long long int k2 = 1;
  long long int c = 1;
  for (int i = 0; i < n - 1; i++) {
    long long int t = i, cnt = 1;
    bool bol = false;
    while (i < n && x[t] == x[i + 1]) {
      bol = true;
      cnt++;
      i++;
    }
    if (bol) {
      i--;
    }
    k1 *= f(cnt);
    k1 = k1 % mod;
  }
  for (int i = 0; i < n - 1; i++) {
    long long int t = i, cnt = 1;
    bool bol = false;
    while (i < n && y[t] == y[i + 1]) {
      bol = true;
      cnt++;
      i++;
    }
    if (bol) {
      i--;
    }
    k2 *= f(cnt);
    k2 = k2 % mod;
  }
  bool fl = false;
  for (int i = 1; i < n; i++) {
    if (p[i].second < p[i - 1].second) {
      fl = true;
    }
  }
  if (fl) {
    c = 0;
  } else {
    for (int i = 0; i < n - 1; i++) {
      long long int t = i, cnt = 1;
      bool bol = false;
      while (i < n && p[t].first == p[i + 1].first &&
             p[t].second == p[i + 1].second) {
        bol = true;
        cnt++;
        i++;
      }
      if (bol) {
        i--;
      }
      c *= f(cnt);
      c = c % mod;
    }
  }
  cout << ((f(n) % mod - (k1 + k2) % mod + mod) % mod + c % mod) % mod << endl;
  return 0;
}
