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.lang.*;
import java.util.*;

public class D1207 {
    public static void main(String[] args) throws IOException{
        long ans;
        StringTokenizer st;
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        st = new StringTokenizer(f.readLine());
        int n = Integer.parseInt(st.nextToken());
        long[] arr1 = new long[n];
        point[] arr = new point[n];
        long MOD = 998244353;
        for(int i = 0; i < n; i++){
            st = new StringTokenizer(f.readLine());
            long ft = Long.parseLong(st.nextToken());
            long sc = Long.parseLong(st.nextToken());
            arr[i] = new point(ft, sc);
            arr1[i] = sc;
        }
        Arrays.sort(arr);
        Arrays.sort(arr1);
        f.close();
        //System.out.println(Arrays.toString(arr));
        //System.out.println(Arrays.toString(arr1));
        long factorial = 1;
        long curr = 1;
        long curr1 = 1;
        long curr2 = 1;
        long f1 = 1;
        long f2 = 1;
        long f3 = 1;
        boolean isSorted = true;
        for(int i = 1; i < n; i++){
            factorial*=(i+1);
            factorial%=MOD;
            if(arr[i].x == arr[i-1].x){
                curr++;
            }else curr = 1;
            if(arr1[i] == arr1[i-1]){
                curr1++;
            }else curr1 = 1;
            if(arr[i].equals(arr[i-1])){
                curr2++;
            }else curr2 = 1;

            if(arr[i].y < arr[i-1].y)
                isSorted = false;
            f1*=curr;
            f2*=curr1;
            f3*=curr2;
            f1%=MOD;
            f2%=MOD;
            f3%=MOD;
        }
        //System.out.println(f1 + " " + f2 + " " + f3 + " " + factorial);
        //System.out.println(f3);
        ans = factorial-f1-f2;
//        if(n == 300000){
//            System.out.println(f1 + " " + f2 + " " + f3 + " " + factorial);
//        }
        if(isSorted)
            ans+=(f3);

        while(ans < 0){
            ans+=MOD;
        }

        System.out.println(ans);
    }
    public static class point implements Comparable<point>{
        long x;
        long y;
        public point(long x,long y){
            this.x = x;
            this.y = y;
        }
        public String toString(){
            return(x + " " + y);
        }
        public boolean equals(Object x){
            point y = ((point)(x));
            if (this.x == y.x && this.y == y.y){
                return true;
            }
            return false;
        }
        public int hashCode(){
            return Objects.hash(x,y);
        }
        public int compareTo(point other){
            if(this.x < other.x || (this.x == other.x && this.y < other.y)){
                return -1;
            }else if(this.x == other.x && this.y == other.y){
                return 0;
            }
            return 1;
        }
    }

}