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;
long long int mod = 998244353;
bool possible = true;
int main() {
  ios::sync_with_stdio(false);
  long long int n, ans = 0, p;
  cin >> n;
  vector<pair<int, int>> a(n, {0, 0});
  unordered_map<long long int, long long int> cnt1, cnt2, cnt3;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first >> a[i].second;
    cnt1[a[i].first]++;
    cnt2[a[i].second]++;
    cnt3[mod * a[i].first + a[i].second]++;
  }
  vector<long long int> fact(n + 1, 0);
  fact[0] = fact[1] = 1;
  for (long long int i = 2; i <= n; i++) {
    fact[i] = (fact[i - 1] * i) % mod;
  }
  p = 1;
  for (auto xx : cnt1) {
    long long int x = xx.second;
    p = (p * fact[x]) % mod;
  }
  ans = (ans + p) % mod;
  p = 1;
  for (auto xx : cnt2) {
    long long int x = xx.second;
    p = (p * fact[x]) % mod;
  }
  ans = (ans + p) % mod;
  sort((a).begin(), (a).end());
  if (possible) {
    for (int i = 1; i < n; i++) {
      long long int x = a[i].first, y = a[i].second;
      long long int x1 = a[i - 1].first, y1 = a[i - 1].second;
      if (x < x1 || y < y1) goto jump;
    }
    p = 1;
    for (auto xx : cnt3) {
      long long int x = xx.second;
      p = (p * fact[x]) % mod;
    }
    ans -= p;
    if (ans < 0) ans += mod;
  }
jump:
  cout << (fact[n] - ans + mod) % mod << '\n';
  return 0;
}
