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;
const int N = int(3e5) + 1;
bool comp(pair<int, int> p1, pair<int, int> p2) {
  if (p1.second != p2.second)
    return p1.second < p2.second;
  else
    return p1.first < p2.first;
}
int sum(int a, int b) { return (a + b) % (int)998244353; }
int mul(int a, int b) { return (a * 1LL * b) % (int)998244353; }
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  int fact[300001];
  fact[0] = 1;
  for (int i = 1; i < N; ++i) fact[i] = mul(i, fact[i - 1]);
  vector<pair<int, int>> v;
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    v.push_back(make_pair(a, b));
  }
  int ans = 0;
  vector<pair<int, int>> vv(v.begin(), v.end());
  ans = 0;
  for (int c = 0; c < 2; c++) {
    sort(v.begin(), v.end());
    int tmp = 1;
    int l = 0;
    while (l < n) {
      int r = l + 1;
      while (r < n && v[l].first == v[r].first) ++r;
      tmp = mul(tmp, fact[r - l]);
      l = r;
    }
    ans = (ans + tmp) % (int)998244353;
    for (int k = 0; k < n; k++) swap(v[k].first, v[k].second);
  }
  vector<int> a, b;
  for (int i = 0; i < n; i++) {
    a.push_back(v[i].first);
    b.push_back(v[i].second);
  }
  if (is_sorted(a.begin(), a.end()) && is_sorted(b.begin(), b.end())) {
    long long tmp = 1;
    for (int i = 0; i < n; i++) {
      int len = 1;
      while (i + 1 < n && v[i] == v[i + 1]) {
        len++;
        i++;
      }
      tmp = tmp * fact[len] % (int)998244353;
    }
    ans = (ans + (int)998244353 - tmp) % (int)998244353;
  }
  cout << (fact[n] - ans + (int)998244353) % (int)998244353;
  return 0;
}
