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 long mod=998244353;
	public static void main(String[] args) throws Exception{
		PrintWriter pw=new PrintWriter(System.out);
		MScanner sc = new MScanner(System.in);
		int n=sc.nextInt();
		int[][]in=new int[n][2];
		for(int i=0;i<n;i++) {
			in[i]=sc.intArr(2);
		}
		Arrays.sort(in,(x,y)->x[0]==y[0]?x[1]-y[1]:x[0]-y[0]);
		
		long[]fac=new long[n+7];
		fac[0]=1;
		for(int i=1;i<fac.length;i++)fac[i]=(fac[i-1]*i)%mod;
		
		long ans=fac[n];
		long remove1=1,remove2=1,removeOverlap=1;
		
		int[][]occ=new int[2][n+1];
		for(int i=0;i<n;i++) {
			for(int j=0;j<2;j++) {
				occ[j][in[i][j]]++;
			}
		}
		for(int i=1;i<=n;i++) {
			remove1=(remove1*fac[occ[0][i]])%mod;
			remove2=(remove2*fac[occ[1][i]])%mod;
		}
		for(int i=1;i<n;i++) {
			if(in[i][0]<in[i-1][0] || in[i][1]<in[i-1][1]) {
				removeOverlap=0;
			}
		}
		int i=0;
		while(i<n) {
			int x=in[i][0],y=in[i][1];
			int cnt=0;
			while(i<n && in[i][0]==x && in[i][1]==y) {
				cnt++;
				i++;
			}
			removeOverlap=(removeOverlap*fac[cnt])%mod;
		}
//		System.out.println(ans+" "+remove1+" "+remove2+" "+removeOverlap);
		ans=(ans-remove1-remove2+removeOverlap)%mod;
		ans=(ans+mod)%mod;
		pw.println(ans);
		
		pw.flush();
	}
	static class MScanner {
		StringTokenizer st;
		BufferedReader br;
		public MScanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
 
		public MScanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}
 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public int[] intArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public long[] longArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public int[] intSortedArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		public long[] longSortedArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		static void shuffle(int[]in) {
			for(int i=0;i<in.length;i++) {
				int idx=(int)(Math.random()*in.length);
				int tmp=in[i];
				in[i]=in[idx];
				in[idx]=tmp;
			}
		}
		static void shuffle(long[]in) {
			for(int i=0;i<in.length;i++) {
				int idx=(int)(Math.random()*in.length);
				long tmp=in[i];
				in[i]=in[idx];
				in[idx]=tmp;
			}
		}
		public Integer[] IntegerArr(int n) throws IOException {
	        Integer[]in=new Integer[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public Long[] LongArr(int n) throws IOException {
	        Long[]in=new Long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public String nextLine() throws IOException {
			return br.readLine();
		}
 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
 
		public char nextChar() throws IOException {
			return next().charAt(0);
		}
 
		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}
 
		public boolean ready() throws IOException {
			return br.ready();
		}
 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
	static void addX(int[]in,int x) {
		for(int i=0;i<in.length;i++)in[i]+=x;
	}
	static void addX(long[]in,int x) {
		for(int i=0;i<in.length;i++)in[i]+=x;
	}
}