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 org.omg.PortableServer.POA;

import java.awt.*;
import java.io.*;
import java.util.*;

public class Abc {
    public static void main(String[] args) throws Exception {
        FastReader sc = new FastReader();
        int n=sc.nextInt();
        long fact[]=new long[n+1];
        long mod=998244353;
        fact[0]=1;
        for (int i=1;i<=n;i++){
            fact[i]=(i*fact[i-1])%mod;
        }
        Point p[]=new Point[n];
        int cnta[]=new int[n];
        int cntb[]=new int[n];
        for (int i=0;i<n;i++){
            int a=sc.nextInt()-1,b=sc.nextInt()-1;
            cnta[a]++;cntb[b]++;
            p[i]=new Point(a,b);
        }
        long ans=fact[n];
        long x=1;
        for (int i=0;i<n;i++){
            x=(x*fact[cnta[i]])%mod;
        }
        ans=(ans-x+mod)%mod;
        x=1;
        for (int i=0;i<n;i++){
            x=(x*fact[cntb[i]])%mod;
        }
        ans=(ans-x+mod)%mod;
        Arrays.sort(p, new Comparator<Point>() {
            @Override
            public int compare(Point o1, Point o2) {
                return o1.x-o2.x;
            }
        });
        if (isSorted(p,false)){
            long cnt=1;
            int c=1;
            for (int i=0;i+1<n;i++){
                if (p[i].x==p[i+1].x && p[i].y==p[i+1].y){
                    c++;
                }else {
                    cnt=(fact[c]*cnt)%mod;
                    c=1;
                }
            }
            cnt=(fact[c]*cnt)%mod;
            ans=(ans+cnt)%mod;
            System.out.println(ans);
            return;
        }
        Arrays.sort(p, new Comparator<Point>() {
            @Override
            public int compare(Point o1, Point o2) {
                return o1.y-o2.y;
            }
        });
        if (isSorted(p,true)){
            long cnt=1;
            int c=1;
            for (int i=0;i+1<n;i++){
                if (p[i].x==p[i+1].x && p[i].y==p[i+1].y){
                    c++;
                }else {
                    cnt=(fact[c]*cnt)%mod;
                    c=1;
                }
            }
            cnt=(fact[c]*cnt)%mod;
            ans=(ans+cnt)%mod;
            System.out.println(ans);
            return;
        }
        System.out.println(ans);


    }
    static boolean isSorted(Point p[],boolean x){
        if (x){
            for (int i=0;i+1<p.length;i++){
                if (p[i].x>p[i+1].x)return false;
            }
        }else {
            for (int i=0;i+1<p.length;i++){
                if (p[i].y>p[i+1].y)return false;
            }
        }
        return true;
    }
    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;
        }
    }
}