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 sum(long long a, long long b) {
  return (a % 998244353 + b % 998244353) % 998244353;
}
long long mul(long long a, long long b) {
  return (a % 998244353 * b % 998244353) % 998244353;
}
long long st(long long a, long long b) {
  return ((a % 998244353 - b % 998244353) + 998244353) % 998244353;
}
int32_t main() {
  cin.sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  long long fact[n + 1];
  fact[0] = 1;
  for (long long i = 1; i <= n; i++) {
    fact[i] = mul(fact[i - 1], i);
  }
  pair<long long, long long> s[n + 3];
  pair<long long, long long> t[n + 3];
  for (long long i = 1; i <= n; i++) {
    cin >> s[i].first >> s[i].second;
    t[i].first = s[i].second;
    t[i].second = s[i].first;
  }
  s[n + 1].first = (long long)(1e17);
  long long f = 1;
  sort(s + 1, s + n + 1);
  vector<vector<long long>> V;
  for (long long i = 1; i <= n;) {
    long long st = i;
    long long el = s[i].first;
    vector<long long> v;
    while (s[i].first == el) {
      v.push_back(s[i].second);
      i++;
      if (i == n + 1) {
        break;
      }
    }
    V.push_back(v);
    f = mul(f, fact[i - st]);
  }
  sort(t + 1, t + n + 1);
  long long g = 1;
  for (long long i = 1; i <= n;) {
    long long st = i;
    long long el = t[i].first;
    while (t[i].first == el) {
      i++;
      if (i == n + 1) {
        break;
      }
    }
    g = mul(g, fact[i - st]);
  }
  long long h = 1;
  for (auto v : V) {
    sort(v.begin(), v.end());
  }
  for (auto v : V) {
    for (long long i = 0; i < v.size();) {
      long long st = i;
      long long el = v[i];
      while (v[i] == el) {
        i++;
        if (i == v.size()) {
          break;
        }
      }
      h = mul(h, fact[i - st]);
    }
  }
  long long sub;
  sub = sum(f, g);
  vector<long long> ch;
  for (auto v : V) {
    for (auto l : v) {
      ch.push_back(l);
    }
  }
  long long fk = 0;
  for (long long i = 0; i < n - 1; i++) {
    if (ch[i] > ch[i + 1]) {
      fk = 1;
      break;
    }
  }
  if (!fk) {
    sub = st(sub, h);
  }
  cout << st(fact[n], sub);
}
