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 long long p = 998244353;
const long long N = 3e5;
long long fact[N + 1];
int32_t main() {
  clock_t clk = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  fact[0] = 1;
  for (long long i = 1; i <= N; i++) {
    fact[i] = (fact[i - 1] * i) % p;
  }
  long long n;
  cin >> n;
  vector<pair<long long, long long> > a(n);
  for (auto &it : a) cin >> it.first >> it.second;
  map<long long, vector<long long> > byA;
  map<long long, vector<long long> > byB;
  long long ans = 0;
  long long temp;
  for (pair<long long, long long> pa : a) {
    byA[pa.first].push_back(pa.second);
  }
  temp = 1;
  for (auto pa : byA) {
    temp = (temp * fact[pa.second.size()]) % p;
  }
  ans = (ans + temp) % p;
  for (pair<long long, long long> pa : a) {
    byB[pa.second].push_back(pa.first);
  }
  temp = 1;
  for (auto pa : byB) {
    temp = (temp * fact[pa.second.size()]) % p;
  }
  ans = (ans + temp) % p;
  bool isPossible = 1;
  for (auto it = byA.begin(); it != byA.end(); it++) {
    sort(it->second.begin(), it->second.end());
    if (it == byA.begin()) continue;
    auto it2 = it;
    --it2;
    long long v1 = it2->second.back();
    long long v2 = it->second[0];
    if (v2 < v1) {
      isPossible = 0;
      break;
    }
  }
  temp = 0;
  if (isPossible) {
    temp = 1;
    for (auto pa : byA) {
      long long m = pa.second.size();
      vector<long long> isLand;
      long long c = 1;
      long long flag = pa.second[0];
      for (long long i = 1; i < m; i++) {
        if (flag == pa.second[i])
          c++;
        else {
          isLand.push_back(c);
          c = 1;
          flag = pa.second[i];
        }
      }
      if (c) isLand.push_back(c);
      long long temp3 = 1;
      for (long long x : isLand) {
        temp3 = (temp3 * fact[x]) % p;
      }
      temp = (temp * temp3) % p;
    }
    ans = (ans + p - temp) % p;
  }
  long long total = fact[n];
  total = (total + p - ans) % p;
  cout << total;
  return 0;
}
