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.*;
public class Main
{
    static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    }
    static final long mod=998244353;
    static class comp implements Comparator<int []>
    {
        public int compare(int a1[],int a2[])
        {
            if(a1[0]!=a2[0])
            return a1[0]-a2[0];
            return a1[1]-a2[1];
        }
    }
    public static void main(String args[])throws Exception
    {
        FastReader fs=new FastReader();
        PrintWriter pw=new PrintWriter(System.out);
        int n=fs.nextInt();
        int arr[][]=new int[n][2];
        for(int i=0;i<n;i++)
        {
            arr[i][0]=fs.nextInt();
            arr[i][1]=fs.nextInt();
        }
        long ans1=1,ans2=1,ans3=1;
        long fac[]=new long[n+2];
        fac[0]=1;
        for(int i=1;i<=n;i++)
        fac[i]=(fac[i-1]*i)%mod;
        Arrays.sort(arr,new comp());
        long ans=0;
        for(int i=0;i<n;i++)
        {
            int j,cnt=1;
            for(j=i+1;j<n;j++)
            {
                if(arr[j][0]!=arr[i][0])
                break;
                cnt++;
            }
            ans1=ans1*fac[cnt];
            ans1%=mod;
            i=j-1;
        }
        for(int i=1;i<n;i++)
        {
            if(arr[i][1]<arr[i-1][1])
            ans3=0;
        }
        Arrays.sort(arr,(a1,a2)->a1[1]-a2[1]);
        for(int i=0;i<n;i++)
        {
            int j,cnt=1;
            for(j=i+1;j<n;j++)
            {
                if(arr[j][1]!=arr[i][1])
                break;
                cnt++;
            }
            ans2=ans2*fac[cnt];
            ans2%=mod;
            i=j-1;
        }
        for(int i=0;i<n;i++)
        {
            int j,cnt=1;
            for(j=i+1;j<n;j++)
            {
                if(arr[j][0]==arr[i][0]&&arr[j][1]==arr[i][1])
                cnt++;
                else
                break;
            }
            i=j-1;
            ans3=ans3*fac[cnt];
            ans3%=mod;
        }
        ans=(ans+ans1)%mod;
        ans=(ans+ans2)%mod;
        ans=(ans-ans3+mod)%mod;
        pw.println((fac[n]-ans+mod)%mod);
        pw.flush();
        pw.close();
    }
}