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 = 1e6 + 300;
const int SQRTN = 320;
const int LOGN = 20;
const double PI = acos(-1);
const double TAU = 2 * PI;
int a[2], n;
int freq[2][N];
map<pair<int, int>, int> m;
int fact[N];
vector<pair<int, int> > v;
int f(int x) {
  if (!x) return 1;
  if (fact[x]) return fact[x];
  return fact[x] = f(x - 1) * 1ll * x % 998244353;
}
bool c() {
  sort(v.begin(), v.end());
  for (int i = 1; i < v.size(); i++)
    if (v[i].second < v[i - 1].second) return false;
  return true;
  ;
}
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int x = 0; x < 2; x++) {
      cin >> a[x];
      freq[x][a[x]]++;
    }
    v.push_back(make_pair(a[0], a[1]));
    m[make_pair(a[0], a[1])]++;
  }
  int ans = f(n);
  int t[2] = {1, 1};
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j < 2; j++) {
      t[j] = t[j] * 1ll * f(freq[j][i]) % 998244353;
    }
  }
  t[0] += t[1];
  t[0] %= 998244353;
  ans -= t[0];
  ans %= 998244353;
  ans += 998244353;
  ans %= 998244353;
  map<pair<int, int>, int>::iterator it = m.begin();
  int aa = 1;
  for (; it != m.end(); it++) {
    aa = aa * 1ll * f(it->second) % 998244353;
  }
  if (!c()) aa = 0;
  ans += aa;
  ans %= 998244353;
  cout << ans << endl;
}
signed main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
