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.*;
import java.util.*;

public class Main
{
    static final int MOD = 998244353 ;
    static long [] fact ;

    static boolean isBoth(Pair [] p)
    {
        Arrays.sort(p , (p1 , p2)-> p1.x != p2.x ? p1.x - p2.x : p1.y - p2.y);
        boolean can = true ;
        for(int i = 1; i < p.length ; i++)
            can &= (p[i - 1].x <= p[i].x && p[i - 1].y <= p[i].y) ;
        if(can)return can ;
        Arrays.sort(p , (p1 , p2)-> p1.y != p2.y ? p1.y - p2.y : p1.x - p2.x);
        can = true ;
        for(int i = 1; i < p.length ; i++)
            can &= (p[i - 1].x <= p[i].x && p[i - 1].y <= p[i].y) ;
        return can ;
    }
    static long sortBoth(Pair [] p)
    {
        TreeMap<Pair , Integer> map = new TreeMap<>(((p1 , p2)-> p1.x != p2.x ? p1.x - p2.x : p1.y - p2.y)) ;
        for(Pair pp : p)
            map.put(pp , map.getOrDefault(pp , 0) + 1) ;
        long ans = 1 ;
        for(int cnt : map.values())
            ans = ans * fact[cnt] % MOD ;
        return ans ;

    }
    static long sortFirst(Pair [] p)
    {
        Arrays.sort(p , (p1 , p2)-> p1.x != p2.x ? p1.x - p2.x : p1.y - p2.y);

        int cnt = 1 ;
        long ans = 1 ;
        for(int i = 1 ; i < p.length ; i ++)
            if(p[i- 1].x == p[i].x)
                cnt ++ ;
            else
            {
                ans = ans  * fact[cnt] % MOD ;
                cnt = 1 ;
            }
        return ans  * fact[cnt] % MOD ;
    }

    static long sortLast(Pair [] p)
    {
        Arrays.sort(p , (p1 , p2)-> p1.y != p2.y ? p1.y - p2.y : p1.x - p2.x);

        int cnt = 1 ;
        long ans = 1 ;
        for(int i = 1 ; i < p.length ; i ++)
            if (p[i - 1].y == p[i].y)
                cnt++;
             else {
                ans = ans * fact[cnt] % MOD;
                cnt = 1;
            }
        return ans  * fact[cnt] % MOD ;
    }


    public static void main (String [] args) throws Exception
    {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out) ;

        int n = sc.nextInt() ;
        Pair [] p = new Pair[n] ;
        fact = new long [n + 1] ;
        fact[0] = 1 ;
        for(int i = 1 ; i <= n ; i++)
            fact[i] = fact[i - 1] * i % MOD ;
        for(int i  = 0; i < n ; i++)
            p[i] = new Pair(sc.nextInt() , sc.nextInt()) ;

        long ans = fact[n] - sortFirst(p.clone()) - sortLast(p.clone());
        if(isBoth(p.clone()))
            ans += sortBoth(p.clone()) ;
        ans %= MOD ;
        ans += MOD ;
        ans %= MOD ;
        out.println(ans);

        out.flush();
    }
    static class Pair
    {
        int x ,   y ;

        Pair(int a , int b)
        {
            x = a ;
            y = b ;

        }
    }
    static class Scanner
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in)) ;
        StringTokenizer st ;

        Scanner(InputStream in) {}

        String next() throws Exception
        {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine()) ;
            return st.nextToken() ;
        }
        int nextInt() throws Exception
        {
            return Integer.parseInt(next()) ;
        }
        double nextDouble() throws Exception
        {
            return Double.parseDouble(next()) ;
        }
        long nextLong() throws Exception
        {
            return Long.parseLong(next()) ;
        }
    }

}