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.awt.*;
import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.PrintStream;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class CF_1207D {
    static long mod = 998244353;

    public static void main(String args[]) {
        Scanner in = new Scanner(new BufferedInputStream(System.in));
        PrintStream out = new PrintStream(new BufferedOutputStream(System.out));

        int n = in.nextInt();
        ArrayList<Point> pts = new ArrayList<>();
        for (int i = 0; i < n; i++) pts.add(new Point(in.nextInt(), in.nextInt()));

        long[] fact = new long[n+1];
        long tot = 1;
        for(int i = 1; i <= n; i++){
            tot = (tot * i) % mod;
            fact[i] = tot;
        }

        ArrayList<Integer> x = new ArrayList<>();
        ArrayList<Integer> y = new ArrayList<>();
        for (Point p : pts) {
            x.add(p.x);
            y.add(p.y);
        }

        Collections.shuffle(x);
        Collections.shuffle(y);
        Collections.shuffle(pts);

        Collections.sort(x);
        Collections.sort(y);
        Collections.sort(pts, (a, b) -> {
            if (a.x == b.x) return a.y - b.y;
            return a.x - b.x;
        });

        long ret = 1;
        for (int i = 0; i < n; ) {
            int curr = x.get(i);
            int prevI = i;
            i++;
            while (i < n && x.get(i) == curr) i++;

            ret = (ret * fact[i - prevI]) % mod;
        }
        long ret2 = 1;
        for (int i = 0; i < n; ) {
            int curr = y.get(i);
            int prevI = i;
            i++;
            while (i < n && y.get(i) == curr) i++;

            ret2 = (ret2 * fact[i - prevI]) % mod;
        }

        int currY = Integer.MIN_VALUE;
        boolean valid = true;
        for(int i = 0; i < n; i++){
            if(pts.get(i).y < currY) {
                valid = false;
                break;
            }
            currY = pts.get(i).y;
        }
        long ret3 = 0;
        if(valid){
            ret3 = 1;
            for(int i = 0; i < n; ){
                Point p = pts.get(i);
                int prevI = i;

                i++;
                while(i < n && pts.get(i).equals(p)) i++;

                ret3 = (ret3 * fact[i - prevI]) % mod;
            }
        }




        long fin = tot - (ret + ret2 - ret3);
        fin = Math.floorMod(fin, mod);

        out.println(fin);


        out.close();
        in.close();
    }
}
