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;
int INF = INT_MAX;
long long mod = 998244353;
long long power(long long a, long long b) {
  long long result = 1;
  while (b > 0) {
    if (b % 2 == 1) result = (result * a) % mod;
    a = (a * a) % mod;
    b /= 2;
  }
  return result;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<pair<pair<int, int>, int>> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i].first.first >> a[i].first.second;
    a[i].second = i;
  }
  sort(a.rbegin(), a.rend());
  vector<pair<int, int>> first(n);
  vector<pair<int, int>> second(n);
  int f;
  int l, r;
  r = 0;
  while (r < n) {
    f = a[r].first.first;
    l = r;
    while (r < n && a[r].first.first == f) {
      r++;
    }
    for (int i = l; i < r; i++) {
      first[a[i].second] = {l, r - 1};
    }
  }
  for (int i = 0; i < n; i++) swap(a[i].first.first, a[i].first.second);
  sort(a.rbegin(), a.rend());
  r = 0;
  while (r < n) {
    f = a[r].first.first;
    l = r;
    while (r < n && a[r].first.first == f) {
      r++;
    }
    for (int i = l; i < r; i++) {
      second[a[i].second] = {l, r - 1};
    }
  }
  long long by_first = 0;
  vector<long long> fact(n + 1);
  fact[0] = 1;
  fact[1] = 1;
  for (int i = 2; i <= n; i++) {
    fact[i] = (fact[i - 1] * i) % mod;
  }
  set<pair<int, int>> dif;
  for (int i = 0; i < n; i++) dif.insert(first[i]);
  set<pair<int, int>>::iterator it;
  it = dif.begin();
  by_first = 1;
  while (it != dif.end()) {
    by_first = (by_first * fact[(*it).second - (*it).first + 1]) % mod;
    it++;
  }
  long long by_second = 1;
  dif.clear();
  for (int i = 0; i < n; i++) dif.insert(second[i]);
  it = dif.begin();
  while (it != dif.end()) {
    by_second = (by_second * fact[(*it).second - (*it).first + 1]) % mod;
    it++;
  }
  bool have = true;
  for (int i = 0; i < n; i++) {
    first[i] = {max(first[i].first, second[i].first),
                min(second[i].second, first[i].second)};
    if (first[i].first > first[i].second) have = false;
  }
  long long intersect = 0;
  if (have) {
    intersect = 1;
    dif.clear();
    for (int i = 0; i < n; i++) dif.insert(first[i]);
    it = dif.begin();
    while (it != dif.end()) {
      intersect = (intersect * fact[(*it).second - (*it).first + 1]) % mod;
      it++;
    }
  }
  long long ans = (by_first + by_second - intersect + mod) % mod;
  ans = (fact[n] - ans + mod) % mod;
  cout << ans << "\n";
}
