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;
long long mul(long long a, long long b) {
  return ((a % 998244353) * (b % 998244353)) % 998244353;
}
long long add(long long a, long long b) {
  return ((a % 998244353) + (b % 998244353)) % 998244353;
}
long long sub(long long a, long long b) {
  return ((a % 998244353) - (b % 998244353)) % 998244353;
}
long long fac(long long x) {
  long long resp = 1;
  for (long long i = 1; i <= x; i++) resp = mul(resp, i);
  return resp;
}
bool cmp(pair<int, int> p1, pair<int, int> p2) {
  if (p1.first < p2.first)
    return 1;
  else if (p1.first > p2.first)
    return 0;
  else
    return p1.second < p2.second;
}
bool cmp1(pair<int, int> p1, pair<int, int> p2) {
  return p1.second < p2.second;
}
pair<int, int> arr[300010];
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i].first >> arr[i].second;
  }
  long long ans = fac(n);
  sort(arr, arr + n);
  long long qtd = 1;
  long long c1 = 1;
  for (int i = 1; i < n; i++) {
    if (arr[i].first == arr[i - 1].first)
      qtd++;
    else {
      c1 = mul(c1, fac(qtd));
      qtd = 1;
    }
  }
  c1 = mul(c1, fac(qtd));
  sort(arr, arr + n, cmp1);
  qtd = 1;
  long long c2 = 1;
  for (int i = 1; i < n; i++) {
    if (arr[i].second == arr[i - 1].second)
      qtd++;
    else {
      c2 = mul(c2, fac(qtd));
      qtd = 1;
    }
  }
  c2 = mul(c2, fac(qtd));
  sort(arr, arr + n, cmp);
  bool sorted = true;
  long long cnt12 = 1;
  qtd = 1;
  for (int i = 1; i < n; i++) {
    if (arr[i].second < arr[i - 1].second) sorted = false;
    if (arr[i] == arr[i - 1])
      qtd++;
    else {
      cnt12 = mul(cnt12, fac(qtd));
      qtd = 1;
    }
  }
  cnt12 = mul(cnt12, fac(qtd));
  if (!sorted) cnt12 = 0;
  ans = sub(ans, c1);
  ans = add(ans, cnt12);
  ans = sub(ans, c2);
  cout << (ans + 998244353) % 998244353 << endl;
}
