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.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class B {
	
	public static void main(String[] args) {
		FastReader scan = new FastReader();
		PrintWriter out = new PrintWriter(System.out);
		Task solver = new Task();
		int t = 1;
		while(t-->0) solver.solve(1, scan, out);
		out.close();
	}
	
	static class Task {
		static final long MOD = 998244353;
		public void solve(int testNumber, FastReader scan, PrintWriter out) {
			int n = scan.nextInt();
			Pair[][] pairs = new Pair[2][n];
			long[] fact = new long[n+1];
			for(int i = 0; i < n; i++) {
				int a = scan.nextInt(), b = scan.nextInt();
				pairs[0][i] = new Pair(a, b);
				pairs[1][i] = new Pair(b, a);
			}
			Arrays.sort(pairs[0]);
			Arrays.sort(pairs[1]);
			fact[0] = 1;
			for(int i = 1; i <= n; i++) fact[i] = mult(fact[i-1], i);
			long[] calcPerm = {1, 1};
			for(int k = 0; k < 2; k++) {		
				for(int i = 0; i < n; i++) {
					int pre = i;
					while(i < n && pairs[k][pre].a == pairs[k][i].a) i++;
					calcPerm[k] = mult(calcPerm[k], fact[i-pre]);
					i--;
				}
			}
			long totalPerm = sub(fact[n], add(calcPerm[0], calcPerm[1]));
			boolean possibleOverlap = true;
			long addPerm = 1;
			int duplicate = 1;
			for(int i = 1; i < n; i++) {;
				if(pairs[0][i].b < pairs[0][i-1].b) possibleOverlap = false;
				if(pairs[0][i].a == pairs[0][i-1].a && pairs[0][i].b == pairs[0][i-1].b) duplicate++;
				else {
					addPerm = mult(addPerm, fact[duplicate]);
					duplicate = 1;
				}
			}
			addPerm = mult(addPerm, fact[duplicate]);
			if(possibleOverlap) totalPerm = add(totalPerm, addPerm);
			out.println(totalPerm);
		}
		
		static long mult(long a, long b) {
			return ((a%MOD)*(b%MOD))%MOD;
		}
		static long add(long a, long b) {
			return (a%MOD+b%MOD)%MOD;
		}
		static long sub(long a, long b) {
			return (a%MOD-b%MOD+MOD)%MOD;
		}
		
		static class Pair implements Comparable<Pair>{
			int a;
			int b;
			public Pair(int aa, int bb) {
				a = aa;
				b = bb;
			}
			
			@Override
			public int compareTo(Pair o) {
				if(a==o.a) return Integer.compare(b, o.b);
				return Integer.compare(a, o.a);
			}
			
		}
	}
	
	static void shuffle(int[] a) {
		Random get = new Random();
		for(int i = 0; i < a.length; i++) {
			int r = get.nextInt(a.length);
			int temp = a[i];
			a[i] = a[r];
			a[r] = temp;
		}
	}
	
	static void shuffle(long[] a) {
		Random get = new Random();
		for(int i = 0; i < a.length; i++) {
			int r = get.nextInt(a.length);
			long temp = a[i];
			a[i] = a[r];
			a[r] = temp;
		}
	}
	

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		public FastReader(String s) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(new File(s)));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

}
