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 gcd(long long a, long long b) {
  if (a > b) swap(a, b);
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long powerMod(long long x, long long y) {
  long long res = 1;
  x %= 998244353;
  while (y > 0) {
    if (y & 1) res = (res * x) % 998244353;
    y = y >> 1;
    x = (x * x) % 998244353;
  }
  return res % 998244353;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n, i, j, k, l;
  cin >> n;
  if (n == 1) return cout << "0\n", 0;
  set<long long> a, b;
  map<long long, long long> mpa, mpb;
  vector<pair<long long, long long> > v(n);
  for (i = 0; i < n; i++) {
    cin >> v[i].first >> v[i].second;
    a.insert(v[i].first);
    mpa[v[i].first]++;
    b.insert(v[i].second);
    mpb[v[i].second]++;
  }
  long long fact[n + 1];
  memset((fact), 0, sizeof((fact)));
  fact[0] = 1;
  for (i = 1; i <= n; i++) {
    fact[i] = (i * fact[i - 1]) % 998244353;
  }
  sort(v.begin(), v.end());
  k = fact[n];
  l = 1;
  for (auto it = a.begin(); it != a.end(); it++) {
    j = *it;
    l = (l * fact[mpa[j]]) % 998244353;
  }
  k = (k - l + 998244353) % 998244353;
  l = 1;
  for (auto it = b.begin(); it != b.end(); it++) {
    j = *it;
    l = (l * fact[mpb[j]]) % 998244353;
  }
  k = (k - l + 998244353) % 998244353;
  for (i = 0; i < n - 1; i++) {
    if (v[i].second > v[i + 1].second) return cout << k << "\n", 0;
  }
  map<pair<long long, long long>, long long> mp;
  for (pair<long long, long long> p : v) mp[p]++;
  l = 1;
  for (auto it = mp.begin(); it != mp.end(); it++) {
    l = (l * fact[it->second]) % 998244353;
  }
  k = (k + l) % 998244353;
  cout << k << "\n";
  return 0;
}
