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

public class Main {
    static Scanner sc = new Scanner(System.in);
    static PrintWriter out = new PrintWriter(System.out);
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    long mod = 998244353;
    long[] fac;
    public static void main(String[] args) throws Exception {
        Main main = new Main();
        // int t = sc.nextInt();
        int t= 1;
        while(t-->0){
            main.solve();
        }
        out.flush();
    }

    void solve() throws Exception {
        int n = sc.nextInt();
        ini(n);
        int[][] arr = new int[n][];
        for(int i=0; i<n; i++){
            arr[i] = new int[]{sc.nextInt(), sc.nextInt()};
        }
        long ans1 = helper(n, arr);
        for(int[] e:arr){
            int mid = e[0]; e[0] = e[1]; e[1] = mid;
        }
        long ans2 = helper(n,arr);
        Arrays.sort(arr,(o1, o2)->(o1[0]==o2[0]?o1[1]-o2[1]:o1[0]-o2[0]));
        boolean flag = true;
        for(int i=0; i<n-1&&flag; i++){
            flag &= arr[i][1]<=arr[i+1][1];
        }
        long ans = (ans1+ans2)%mod;
        if(flag){
            long temp = 1;
            int left= 0;
            while(left<n){
                int right = left;
                while(right<n&&(arr[right][0]==arr[left][0]&&arr[right][1]==arr[left][1])){
                    right++;
                }
                temp = temp*fac[right-left]%mod;
                left=right;
            }
            ans = (ans+mod-temp%mod)%mod;
        }
        ans = (fac[n]-ans+mod)%mod;
        out.println(ans);
    }
    long helper(int n, int[][] arr){
        long ans = 1;
        Arrays.sort(arr, Comparator.comparingInt(o1->o1[0]));
        int left = 0;
        while(left<n){
            int right = left;
            while(right<n && arr[right][0] == arr[left][0]){
                right++;
            }
            int len = right-left;
            ans = ans*fac[len]%mod;
            left= right;
        }
        return ans;
    }
    void ini(int n){
        fac = new long[n+1];
        fac[0] = 1;
        for(int i=1; i<=n; i++) fac[i] = fac[i-1]*i%mod;
    }

    int paIn(String s){return Integer.parseInt(s);}

}