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.lang.*;
import java.lang.reflect.Array;
import java.util.*;
import java.io.*;

public class Main {
    void solve() {
        int n=ni();
        int a[]=new int[n+1];
        int b[]=new int[n+1];
        for(int i=1;i<=n;i++){
            a[i]=ni(); b[i]=ni();
        }
        precompute(n);
        long ans=get(a,n)+get(b,n);
        if(ans>=M) ans%=M;

        Pair p[]=new Pair[n+1];
        for(int i=1;i<=n;i++){
            p[i]=new Pair(a[i],b[i]);
        }
        Arrays.sort(p,1,n+1);

        if(check(p,n)){

            long ans2=1;
            int cc,j;
            for(int i=1;i<=n;){
                cc=1;
                j=i+1;
                while(j<=n && p[i].x==p[j].x && p[i].y==p[j].y) {
                    cc++;
                    j++;
                }
                ans2=mul(ans2,fact[cc]);
                i=j;
            }

            ans-=ans2;
            if(ans<0) ans+=M;
        }
        ans=fact[n]-ans;
        if(ans<0) ans+=M;
        pw.println(ans);


    }
    long fact[];
    boolean check(Pair p[],int n){
        for(int i=1;i<n;i++){
            if(!(p[i].x<=p[i+1].x && p[i].y<=p[i+1].y)) return false;
        }
        return true;
    }
    long get(int a[],int n){
        int cc[]=new int[n+1];
        for(int i=1;i<=n;i++) cc[a[i]]++;
        long ans=1;
        for(int i=1;i<=n;i++) ans=mul(ans,fact[cc[i]]);
        return ans;

    }
    void precompute(int n){
        fact=new long[n+1];
        fact[0]=1;
        for(int i=1;i<=n;i++) fact[i]=(fact[i-1]*i)%M;
    }
    long add(long a,long b){
        a+=b;
        if(a>=M )a-=M;
        return a;
    }
    long mul(long a,long b){
        a*=b;
        if(a>=M)a%=M;
        return a;
    }
    class Pair implements Comparable<Pair>{
        int x;
        int y;
        public Pair(int x,int y){
            this.x=x;
            this.y=y;
        }
        public int compareTo(Pair c){
            int t=Integer.compare(this.x,c.x);
            if(t!=0) return t;
            return Integer.compare(this.y,c.y);
        }
        public boolean equals(Object o){
            Pair c=(Pair)o;
            return x==c.x && y==c.y;
        }
        public int hashCode(){
            return x+y;
        }

    }

    long M = 998244353;

    // END
    PrintWriter pw;
    StringTokenizer st;
    BufferedReader br;

    void run() throws Exception {
        br = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(System.out);
        long s = System.currentTimeMillis();
        solve();
        pw.flush();
    }
    public static void main(String[] args) throws Exception {
        new Main().run();
    }

    String ns() {
        while (st == null || !st.hasMoreElements()) {

            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    String nextLine() throws Exception {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            throw new Exception(e.toString());
        }
        return str;
    }

    int ni() {
        return Integer.parseInt(ns());
    }

    long nl() {
        return Long.parseLong(ns());
    }

    double nd() {
        return Double.parseDouble(ns());
    }
}