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 M = 998244353;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  vector<pair<long long, long long> > a(n);
  map<long long, long long> mp1, mp2;
  map<pair<long long, long long>, long long> mp;
  long long f = 1;
  for (long long i = 0; i < n; i++) {
    cin >> a[i].first >> a[i].second;
    mp1[a[i].first]++;
    mp2[a[i].second]++;
    mp[a[i]]++;
    f = (f * (i + 1)) % M;
  }
  long long p1 = 0, p2 = 0, p3 = 0;
  for (auto u : mp1) {
    if (!p1) {
      p1 = 1;
    }
    long long f = u.second;
    while (f--) {
      p1 = (p1 * (f + 1)) % M;
    }
  }
  for (auto u : mp2) {
    if (!p2) {
      p2 = 1;
    }
    long long f = u.second;
    while (f--) {
      p2 = (p2 * (f + 1)) % M;
    }
  }
  for (auto u : mp) {
    if (!p3) {
      p3 = 1;
    }
    long long f = u.second;
    while (f--) {
      p3 = (p3 * (f + 1)) % M;
    }
  }
  sort(a.begin(), a.end());
  for (long long i = 1; i < n; i++) {
    if (a[i].second < a[i - 1].second) {
      p3 = 0;
    }
  }
  cout << (f - p1 - p2 + p3 + 10 * M) % M << endl;
  return 0;
}
