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 D {

	static final long mod = 998244353;  
	static long[] fac = new long[300001];
	public static void main(String[] args) throws Exception {
		fac[0] = 1;
		for(int i = 1; i <= 300000; i++){
			fac[i] = (fac[i-1] * i) % mod;
		}
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		HashMap<Integer, Integer> as = new HashMap<>();
		HashMap<Integer, Integer> bs = new HashMap<>();
		HashMap<Pair, Integer> ps = new HashMap<>();
		TreeMap<Integer, Pair> stats = new TreeMap<>();
		for(int i = 0; i < n; i++){
			st = new StringTokenizer(br.readLine());
			int a = Integer.parseInt(st.nextToken());
			int b = Integer.parseInt(st.nextToken());
			add(as, a); add(bs, b); add(ps, new Pair(a, b));
			if(stats.containsKey(a)){
				Pair p = stats.get(a);
				stats.put(a, new Pair(Integer.min(b, p.a), Integer.max(b, p.b)));
			}
			else{
				stats.put(a, new Pair(b, b));
			}
		}
		long tot = (fac[n] + (mod-count(as)) + (mod-count(bs))) % mod;
		boolean check = true;
		Pair prev = null;
		for(int key: stats.keySet()){
			Pair p = stats.get(key);
			if(prev != null && p.a < prev.b) {
				check = false; break; 
			}
			else{
				prev = p;
			}
		}
		if(check) System.out.println((tot + countp(ps)) % mod);
		else System.out.println(tot % mod);
		
	}
	static void add(HashMap<Integer, Integer> map, int key){
		if(map.containsKey(key)){
			map.put(key, map.get(key)+1);
		}
		else{
			map.put(key, 1);
		}
	}
	static void add(HashMap<Pair, Integer> map, Pair key){
		if(map.containsKey(key)){
			map.put(key, map.get(key)+1);
		}
		else{
			map.put(key, 1);
		}
	}
	static long count(HashMap<Integer, Integer> map){
		long tot = 1;
		for(Object key: map.keySet()){
			tot = (tot * fac[map.get(key)]) % mod;
		}
		return tot;
	}
	static long countp(HashMap<Pair, Integer> map){
		long tot = 1;
		for(Object key: map.keySet()){
			tot = (tot * fac[map.get(key)]) % mod;
		}
		return tot;
	}
	static class Pair implements Comparable<Pair>{
		int a, b;
		public Pair(int a, int b){
			this.a = a; this.b = b;
		}
		@Override
		public int compareTo(Pair p) {
			if(a == p.a) return b - p.b;
			else return a - p.a;
		}
		@Override
		public boolean equals(Object o){
			return compareTo((Pair)o) == 0;
		}
		@Override
		public int hashCode(){
			return a+b;
		}	
		public String toString(){
			return a+" "+b;
		}
	}

}
