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 long long mod = 998244353;
long long fak[400000];
bool cmp1(pair<long long, long long> x, pair<long long, long long> y) {
  return x.first < y.first;
}
bool cmp2(pair<long long, long long> x, pair<long long, long long> y) {
  return x.second < y.second;
}
bool cmp3(pair<long long, long long> x, pair<long long, long long> y) {
  if (x.first == y.first) {
    return x.second < y.second;
  } else {
    return x.first < y.first;
  }
}
pair<long long, long long> nom[500000];
int main() {
  long long fi = 1, se = 1, doub = 1;
  fak[0] = 1;
  for (long long i = 1; i < 400000; i++) {
    fak[i] = i * fak[i - 1];
    fak[i] %= mod;
  }
  int N;
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> nom[i].first >> nom[i].second;
  }
  sort(nom, nom + N, cmp1);
  long long coun = 1;
  for (int i = 1; i < N; i++) {
    if (nom[i].first == nom[i - 1].first) {
      coun++;
    } else {
      fi *= fak[coun];
      fi %= mod;
      coun = 1;
    }
  }
  fi *= fak[coun];
  fi %= mod;
  coun = 1;
  sort(nom, nom + N, cmp2);
  for (int i = 1; i < N; i++) {
    if (nom[i].second == nom[i - 1].second) {
      coun++;
    } else {
      se *= fak[coun];
      se %= mod;
      coun = 1;
    }
  }
  se *= fak[coun];
  se %= mod;
  coun = 1;
  sort(nom, nom + N, cmp3);
  for (int i = 1; i < N; i++) {
    if (nom[i].second < nom[i - 1].second) doub = 0;
  }
  for (int i = 1; i < N; i++) {
    if (nom[i] == nom[i - 1]) {
      coun++;
    } else {
      doub *= fak[coun];
      doub %= mod;
      coun = 1;
    }
  }
  doub *= fak[coun];
  doub %= mod;
  long long ans = fak[N] - (fi + se - doub);
  while (ans < 0) ans += mod;
  cout << ans << '\n';
}
