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.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;
 
public class permutations{
	
	public static void main(String[] args) throws IOException
	{
		new permutations().run();
	}
	
	int MOD = 998244353;
	long[] factMod;
 
	public void run() throws IOException
	{
		BufferedReader file = new BufferedReader(new InputStreamReader(System.in));
		int N = Integer.parseInt(file.readLine());
        factMod = new long[N+1];
        factMod[0] = 1;
        for(int i = 1;i<factMod.length;i++)
        		factMod[i] = factMod[i-1]*i%MOD;
        pair[] pairs = new pair[N];
        int[] occ1 = new int[N+1];
        int[] occ2 = new int[N+1];
		for(int i = 0;i<N;i++) {
			StringTokenizer st = new StringTokenizer(file.readLine());
			int a = Integer.parseInt(st.nextToken());
			int b = Integer.parseInt(st.nextToken());
			occ1[a]++;
			occ2[b]++;
			pairs[i] = new pair(a,b);
		}
		long first = 1;
		for(int i = 0;i<occ1.length;i++)
			if(occ1[i]>1)
			{
				first *= factMod[occ1[i]];
				first %= MOD;
			}
		long second = 1;
		for(int i = 0;i<occ2.length;i++)
		{
			if(occ2[i]>1)
			{
				second *= factMod[occ2[i]];
				second %= MOD;
			}
		}
		long both = 1;
		Arrays.sort(pairs);
		boolean sorted = true;
		for(int i = 1;i<pairs.length;i++)
			if(pairs[i].a < pairs[i-1].a || pairs[i].b < pairs[i-1].b)
				sorted = false;
		int run = 1;
		for(int i = 1;i<pairs.length;i++)
		{
			if(pairs[i].a == pairs[i-1].a && pairs[i].b == pairs[i-1].b)
			{
				run++;
			}else {
				both*=factMod[run];
				both %= MOD;
				run = 1;
			}
		}
		both *= factMod[run];
		both %= MOD;
		if(!sorted)
			both = 0;
		System.out.println((((factMod[N] - first - second + both)%MOD)+MOD)%MOD);
	}
	
	private class pair implements Comparable<pair>{
		
		int a,b;
		
		public pair(int a, int b)
		{
			this.a = a;
			this.b = b;
		}
		
		public int compareTo(pair p)
		{
			if(Integer.compare(a, p.a)!=0)
				return Integer.compare(a, p.a);
			return Integer.compare(b, p.b);
		}
		
	}
}