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. 
import java.util.*;
import java.io.*;
import java.util.HashMap;
import java.util.Map;
public class A
{
      public static void main(String ar[]) throws Exception
      {
            BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
            int n=Integer.parseInt(br.readLine());
            int a[][]=new int[n][2];
            
            int b[]=new int[n];
            int c[]=new int[n];
            
            HashMap<String,Integer> hm=new HashMap<String,Integer>();
            HashMap<Integer,Integer> hm1=new HashMap<Integer,Integer>();
            HashMap<Integer,Integer> hm2=new HashMap<Integer,Integer>();
            for(int i=0;i<n;i++)
            {
                  String s1[]=br.readLine().split(" ");
                  int u=Integer.parseInt(s1[0]);
                  int v=Integer.parseInt(s1[1]);
                  
                  a[i][0]=u;
                  a[i][1]=v;
                  
                  String ss=u+" "+v;
                  if(hm.containsKey(ss))
                   hm.put(ss,1+hm.get(ss));
                  else
                   hm.put(ss,1);
                  
                  if(hm1.containsKey(u))
                   hm1.put(u,1+hm1.get(u));
                  else
                   hm1.put(u,1);
                  
                  if(hm2.containsKey(v))
                   hm2.put(v,1+hm2.get(v));
                  else
                   hm2.put(v,1);
            }
            
            
            Arrays.sort(a,new Comparator<int[]>(){
                  public int compare(final int f1[],final int f2[])
                  {
                        if(f1[0]>f2[0])
                         return 1;
                        else if(f1[0]==f2[0])
                        {
                        if(f1[1]>f2[1])
                         return 1;
                        else if(f1[1]==f2[1])
                         return 0;
                        else
                         return -1;
                        }
                        else
                         return -1;
                  }
            });
            
            long x3=1;
            for(int i=1;i<n;i++)
             if(a[i][0]<a[i-1][0] || a[i][1]<a[i-1][1])
              x3=0;
            
            long mod=998244353;
            long f[]=new long[300001];
            f[0]=1;
            f[1]=1;
            for(int i=2;i<=300000;i++)
             f[i]=(f[i-1]*i)%mod;
            
            long x1=1,x2=1;
            for(int i:hm1.keySet())
             x1=(x1*f[hm1.get(i)])%mod;
            
            for(int i:hm2.keySet())
             x2=(x2*f[hm2.get(i)])%mod;
            
            for(String ss:hm.keySet())
             x3=(x3*f[hm.get(ss)])%mod;
            
            long S=(4*mod+f[n]-(x1+x2-x3))%mod;
            
            System.out.println(S);
      }
}