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 z = 998244353;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (res * a) % z;
      b--;
    } else {
      a = (a * a) % z;
      b = b >> 1;
    }
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long fact[300001];
  fact[0] = 1;
  for (long long i = 1; i <= 300000; i++) {
    fact[i] = i * fact[i - 1];
    fact[i] %= z;
  }
  long long t;
  t = 1;
  while (t--) {
    long long n;
    cin >> n;
    long long a, b;
    vector<pair<long long, long long> > v1, v2;
    for (long long i = 0; i < n; i++) {
      cin >> a >> b;
      v1.push_back(make_pair(a, b));
      v2.push_back(make_pair(b, a));
    }
    sort(v1.begin(), v1.end());
    sort(v2.begin(), v2.end());
    vector<long long> v3, v4;
    long long len = 1;
    for (long long i = 1; i < n; i++) {
      if (v1[i].first == v1[i - 1].first)
        len++;
      else {
        v3.push_back(len);
        len = 1;
      }
    }
    v3.push_back(len);
    len = 1;
    for (long long i = 1; i < n; i++) {
      if (v2[i].first == v2[i - 1].first)
        len++;
      else {
        v4.push_back(len);
        len = 1;
      }
    }
    v4.push_back(len);
    long long ans = fact[n], p = 1, q = 1;
    for (long long i = 0; i < v3.size(); i++) {
      p *= fact[v3[i]];
      p %= z;
      if (i > 0) v3[i] += v3[i - 1];
    }
    for (long long i = 0; i < v4.size(); i++) {
      q *= fact[v4[i]];
      q %= z;
      if (i > 0) v4[i] += v4[i - 1];
    }
    p += q;
    p %= z;
    q = 1;
    len = 1;
    long long c = 0;
    for (long long i = 1; i < n; i++) {
      if (v1[i].first == v1[i - 1].first && v1[i].second == v1[i - 1].second)
        len++;
      else {
        if (v1[i].second < v1[i - 1].second) {
          c++;
          break;
        } else {
          q *= fact[len];
          q %= z;
          len = 1;
        }
      }
    }
    q *= fact[len];
    q %= z;
    if (!c) {
      p -= q;
      if (p < 0) p += z;
    }
    ans -= p;
    if (ans < 0) ans += z;
    cout << ans << "\n";
  }
}
