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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class d {
    static BufferedReader s = new BufferedReader(new InputStreamReader(System.in));
    public static void main(String[] args) throws IOException {
        StringBuilder sb=new StringBuilder();
        String[] s=s();
         int n=i(s[0]);
         Student12[] s12=new Student12[n];
         for(int i=0;i<n;i++){
             String[] s1=s();
             int a=i(s1[0]);int b=i(s1[1]);
             s12[i]=new Student12(a,b);
         }int mod=998244353;
       MAXN=3*100000+3;
        fac=new long[MAXN+1];calculatefac(mod);
       int count=1;long ans2=1;
       Arrays.sort(s12,new Sortbyroll123());
       for(int i=1;i<n;i++){
            if(s12[i].r==s12[i-1].r){
                count++;
                if(i==n-1){
                    ans2=((ans2%mod)*(fac[count]%mod))%mod;
                    count=1;
                }
            }else{
                ans2=((ans2%mod)*(fac[count]%mod))%mod;
                count=1;
            }
        }
        long ans1=1;count=1;
        Arrays.sort(s12,new Sortbyroll12());
         for(int i=1;i<n;i++){
            if(s12[i].l==s12[i-1].l){
                count++;
                if(i==n-1){
                    ans1=((ans1%mod)*(fac[count]%mod))%mod;
                    count=1;
                }
            }else{
                ans1=((ans1%mod)*(fac[count]%mod))%mod;
                count=1;
            }
         }
         int flag=0;

        long both=1;count=1;
        for(int i=1;i<n;i++){
            if(s12[i].r==s12[i-1].r&&s12[i].l==s12[i-1].l){
                count++;
                if(i==n-1){
                    both=((both%mod)*(fac[count]%mod))%mod;
                    count=1;
                }
            }
            else if(s12[i-1].r>s12[i].r){
                both=0;break;
            }
            else{
                both=((both%mod)*(fac[count]%mod))%mod;
                count=1;
            }
        }
//        System.out.println(ans1+" "+ans2+" "+both+" "+fac[n]);
        System.out.println((fac[n]-(ans1+ans2-both)+2*mod)%mod);
    }        static String[] s() throws IOException {
        return s.readLine().trim().split("\\s+");
    }
    static int i(String ss) {
        return Integer.parseInt(ss);
    }
    static long l(String ss) {
        return Long.parseLong(ss);
    }
    static long[] fac ;static int MAXN;
    static void calculatefac(int mod){
        fac[0]=1;
        for (int i = 1 ;i <= MAXN; i++)
            fac[i] = fac[i-1] * i % mod;
    }
}
class Student12 {
    int l;int r;
    public Student12(int l, int r) {
        this.l = l;
        this.r = r;
    }

    public String toString()
    {
        return this.l+" ";
    }
}
class Sortbyroll12 implements Comparator<Student12> {

    public int compare(Student12 a, Student12 b){
        if(a.l==b.l){
            return a.r-b.r;
        }
        return a.l-b.l;
        /*if(a.r<b.r) return 1;
        else if(a.r==b.r){
             return a.l-b.l;
            }
        return -1;*/  }
}
class Sortbyroll123 implements Comparator<Student12> {

    public int compare(Student12 a, Student12 b){
        if(a.r==b.r){
            return a.l-b.l;
        }
        return a.r-b.r;
        /*if(a.r<b.r) return 1;
        else if(a.r==b.r){
             return a.l-b.l;
            }
        return -1;*/  }
}