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;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
int n;
vector<int> nxt[300005];
pair<int, int> p[300005];
long long int fact[300005];
void gen() {
  fact[1] = 1;
  for (long long int i = 2; i <= n; i++) {
    fact[i] = (fact[i - 1] * i) % 998244353;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  gen();
  map<int, int> freq1, freq2;
  for (int i = 0; i < n; i++) {
    cin >> p[i].first >> p[i].second;
    freq1[p[i].first]++;
    freq2[p[i].second]++;
    nxt[p[i].first].push_back(p[i].second);
  }
  long long int ans = fact[n];
  long long int cur = 1ll;
  for (auto x : freq1) cur = (cur * fact[x.second]) % 998244353;
  ans = (998244353 + ans - cur) % 998244353;
  cur = 1ll;
  for (auto x : freq2) cur = (cur * fact[x.second]) % 998244353;
  ans = (998244353 + ans - cur) % 998244353;
  long long int prev = 0;
  long long int add = 1;
  bool prob = false;
  for (int i = 1; i <= n && !prob; i++) {
    map<int, int> mp;
    sort(nxt[i].begin(), nxt[i].end());
    for (auto x : nxt[i]) {
      if (prev > x) prob = true;
      prev = x;
      mp[x]++;
    }
    for (auto x : mp) {
      add = (add * fact[x.second]) % 998244353;
    }
  }
  if (prob) add = 0;
  ans = (ans + add) % 998244353;
  cout << ans;
  return 0;
}
