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 maxn = 3e5;
long long fact[maxn + 10], mod = 998244353;
pair<int, int> niz[maxn + 10];
int n;
map<int, int> mapa1, mapa2;
map<pair<int, int>, int> mapa12;
bool srt(pair<int, int> a, pair<int, int> b) {
  return (a.first < b.first) || (a.first == b.first && a.second < b.second);
}
bool check() {
  sort(niz + 1, niz + n + 1, srt);
  for (int i = 1; i < n; i++)
    if (niz[i].first <= niz[i + 1].first && niz[i].second <= niz[i + 1].second)
      continue;
    else
      return false;
  return true;
}
int main() {
  fact[0] = 1;
  for (int i = 1; i <= maxn; i++) fact[i] = (fact[i - 1] * i) % mod;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    niz[i] = {a, b};
    mapa1[a]++;
    mapa2[b]++;
    mapa12[{a, b}]++;
  }
  long long rez1 = 1;
  for (map<int, int>::iterator it = mapa1.begin(); it != mapa1.end(); it++)
    rez1 = (rez1 * fact[it->second]) % mod;
  long long rez2 = 1;
  for (map<int, int>::iterator it = mapa2.begin(); it != mapa2.end(); it++)
    rez2 = (rez2 * fact[it->second]) % mod;
  long long rez12 = 1;
  for (map<pair<int, int>, int>::iterator it = mapa12.begin();
       it != mapa12.end(); it++)
    rez12 = (rez12 * fact[it->second]) % mod;
  long long rez = 0;
  if (check()) {
    rez = (rez1 + rez2 - rez12) % mod;
    if (rez < 0) rez += mod;
    rez = (fact[n] - rez) % mod;
    if (rez < 0) rez += mod;
    printf("%lld\n", rez);
  } else {
    rez = (rez1 + rez2) % mod;
    rez = (fact[n] - rez) % mod;
    if (rez < 0) rez += mod;
    printf("%lld\n", rez);
  }
  return 0;
}
