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 f[300005];
long long expo(long long base, long long exponent, long long mod) {
  long long ans = 1;
  while (exponent != 0) {
    if ((exponent & 1) == 1) {
      ans = ans * base;
      ans = ans % mod;
    }
    base = base * base;
    base %= mod;
    exponent >>= 1;
  }
  return ans % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, i, j;
  cin >> n;
  f[0] = 1;
  for (i = 1; i <= n; i++) f[i] = (f[i - 1] * i) % 998244353;
  vector<pair<long long int, long long int> > a;
  map<long long, long long> m1, m2;
  map<pair<long long int, long long int>, long long> m3;
  for (i = 0; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    a.push_back({x, y});
    m1[x]++;
    m2[y]++;
    m3[{x, y}]++;
  }
  sort((a).begin(), (a).end());
  long long a1 = 1, a2 = 1, a3 = 1;
  vector<long long int> v1, v2, v3;
  for (auto i : m1)
    if (i.second > 1) v1.push_back(i.second);
  for (auto i : m2)
    if (i.second > 1) v2.push_back(i.second);
  for (auto i : m3)
    if (i.second > 1) v3.push_back(i.second);
  for (auto i : v1) {
    a1 = (a1 * f[i]) % 998244353;
  }
  for (auto i : v2) {
    a2 = (a2 * f[i]) % 998244353;
  }
  long long ff = 0;
  for (i = 1; i < n; i++)
    if (a[i].second < a[i - 1].second) ff = 1;
  long long ans = (f[n] - a1 - a2 + 2 * 998244353) % 998244353;
  if (!ff) {
    for (auto i : v3) {
      a3 = (a3 * f[i]) % 998244353;
    }
    ans += a3;
    ans %= 998244353;
  }
  cout << ans;
  return 0;
}
