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 = 300001;
const int mod = 998244353;
int main() {
  ios_base::sync_with_stdio(false);
  long long int fact[N];
  vector<pair<int, int> > seq;
  map<int, long long int> freqA;
  map<int, long long int> freqB;
  map<pair<int, int>, long long int> freqSeq;
  seq.clear();
  freqA.clear();
  freqB.clear();
  freqSeq.clear();
  fact[0] = fact[1] = 1;
  for (int i = 2; i < N; i++) {
    fact[i] = (i * fact[i - 1]) % mod;
  }
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int ai;
    int bi;
    cin >> ai >> bi;
    freqA[ai]++;
    freqB[bi]++;
    seq.push_back(make_pair(ai, bi));
    freqSeq[make_pair(ai, bi)]++;
  }
  sort(seq.begin(), seq.end());
  bool unsorted = false;
  for (int i = 1; i < n; i++) {
    if (seq[i - 1].second > seq[i].second) {
      unsorted = true;
      break;
    }
  }
  long long int res = fact[n];
  long long int res1 = 1;
  for (map<int, long long int>::iterator it = freqA.begin(); it != freqA.end();
       it++) {
    res1 = (res1 * fact[(it->second)]) % mod;
  }
  long long int res2 = 1;
  for (map<int, long long int>::iterator it = freqB.begin(); it != freqB.end();
       it++) {
    res2 = (res2 * fact[(it->second)]) % mod;
  }
  long long int res3 = 1;
  if (!unsorted) {
    for (map<pair<int, int>, long long int>::iterator it = freqSeq.begin();
         it != freqSeq.end(); it++) {
      res3 = (res3 * fact[(it->second)]) % mod;
    }
  }
  res = (res - res1 - res2 + mod + mod) % mod;
  if (!unsorted) {
    res = (res + res3) % mod;
  }
  cout << res % mod << endl;
  return 0;
}
