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.*;
import java.text.*;

public class D1207 {

	static final int mod = 998244353;

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		int n = sc.nextInt();
		Pair[] a = new Pair[n];
		int[] cntx = new int[n + 1];
		int[] cnty = new int[n + 1];
		TreeMap<Pair,Integer> tm = new TreeMap<Pair,Integer>();
		for (int i = 0; i < a.length; i++) {
			a[i] = new Pair(sc.nextInt(), sc.nextInt());
			tm.put(a[i], tm.getOrDefault(a[i],0)+1);
			cntx[a[i].x]++;
			cnty[a[i].y]++;
		}
		long[] factorial = new long[n + 1];
		factorial[0] = 1;
		for (int i = 1; i < n + 1; i++) {
			factorial[i] = ((factorial[i - 1] % mod) * (i % mod)) % mod;
		}
		long ansx = 1;
		for (int i = 0; i < cntx.length; i++) {
			ansx*=factorial[cntx[i]];
			ansx%=mod;
		}
		long ansy = 1;
		for (int i = 0; i < cnty.length; i++) {
			ansy*=factorial[cnty[i]];
			ansy%=mod;
		}
		Arrays.sort(a);
		boolean b = true;
		for (int i = 0; i < factorial.length-2 && b; i++) {
			if(a[i+1].y <a[i].y)
				b = false;
		}
		if(b) {
			long inter =1;
			while(!tm.isEmpty()) {
				inter *= factorial[tm.pollFirstEntry().getValue()];
				inter%=mod;
			}
//			pw.println(ansx+" "+ansy+" "+inter);
			long ans = (ansx+(ansy-inter+mod)%mod)%mod;
			ans = (factorial[n]-ans) %mod;
			pw.println((ans+mod)%mod);
			
		}else {
			long ans = (factorial[n]-(ansx+ansy)%mod)%mod;
			pw.println((ans+mod)%mod);
		}
		pw.close();
		
		

	}

	static class Pair implements Comparable<Pair>{
		int x, y;

		public Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}

		public int compareTo(Pair p) {
			return x == p.x ? y - p.y : x - p.x;
		}
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(FileReader r) {
			br = new BufferedReader(r);
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

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

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public double nextDouble() throws IOException {
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if (x.charAt(0) == '-') {
				neg = true;
				start++;
			}
			for (int i = start; i < x.length(); i++)
				if (x.charAt(i) == '.') {
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				} else {
					sb.append(x.charAt(i));
					if (dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg ? -1 : 1);
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

	}
}
