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.*;

public class numberof {
	static final int MOD=998244353;
	public static void main(String[] args) {
		
		long[] fac=new long[300001];
		fac[0]=1;
		for(int i=1;i<=300000;i++) fac[i]=(fac[i-1]*i)%MOD;
		Scanner scan=new Scanner(System.in);
		int n=scan.nextInt();
		pair[] a=new pair[n];
		for(int i=0;i<n;i++) a[i]=new pair(scan.nextInt(),scan.nextInt());
		Arrays.sort(a,new sort1());
		
		long ways1=1L;
		
		int at=0;
		while(at<n) {
			int nxt=at;
			while(nxt<n&&a[nxt].x==a[at].x) nxt++;
			ways1=(ways1*fac[nxt-at])%MOD;
			at=nxt;
		}
		
		Arrays.sort(a,new sort2());
		
		long ways2=1L;
		
		at=0;
		while(at<n) {
			int nxt=at;
			while(nxt<n&&a[nxt].y==a[at].y) nxt++;
			ways2=(ways2*fac[nxt-at])%MOD;
			at=nxt;
		}
		
		//and now subtract out the number of ways they can both be sorted
		
		long ways3=1L;
		
		Arrays.sort(a,new sort1());
		
		at=0;
		int min=0;
		while(at<n) {
			int nxt=at;
			while(nxt<n&&a[nxt].x==a[at].x) {
				if(a[nxt].y<min) ways3=0L;
				nxt++;
			}
			min=a[nxt-1].y;
			at=nxt;
		}
		
		//and now check duplicates
		
		at=0;
		while(at<n) {
			int nxt=at;
			while(nxt<n&&a[nxt].y==a[at].y&&a[nxt].x==a[at].x) {
				nxt++;
			}
			ways3=(ways3*fac[nxt-at])%MOD;
			at=nxt;
		}
//		System.out.println(ways1+" "+ways2+" "+ways3);
		System.out.println((fac[n]+2*MOD-(ways1+ways2-ways3+2*MOD)%MOD)%MOD);
	}
	static class pair {
		int x,y;
		pair(int x, int y) {
			this.x=x;
			this.y=y;
		}
	}
	static class sort1 implements Comparator<pair> {
		public int compare(pair a, pair b) {
			if(a.x==b.x) return a.y-b.y; 
			return a.x-b.x;
		}
	}
	static class sort2 implements Comparator<pair> {
		public int compare(pair a, pair b) {
			if(a.y==b.y) return a.x-b.x; 
			return a.y-b.y;
		}
	}
}