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.math.*;
public class Solution{
    public static long mod = 998244353;
    public static long[] fact;
    public static void main(String[] args)throws IOException{
        BufferedReader br =  new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        fact = new long[n+1];
        fact[0] = 1;
        for(int i=1;i<=n;i++){
            fact[i] = fact[i-1]*(long)i;
            fact[i] %= mod;
        }
        long res = 1;
        TreeMap<Integer,Integer> max = new TreeMap<Integer,Integer>();
        TreeMap<Integer,Integer> min = new TreeMap<Integer,Integer>();
        min.put(n+1,Integer.MAX_VALUE);
        int[] a = new int[n];
        int[] b = new int[n];
        for(int i=0;i<n;i++){
            st = new StringTokenizer(br.readLine());
            a[i] = Integer.parseInt(st.nextToken());
            b[i] = Integer.parseInt(st.nextToken());
            int di = b[i];
            int da = b[i];
            if(min.containsKey(a[i])) di = Math.min(di,min.get(a[i]));
            if(max.containsKey(a[i])) da = Math.max(da,max.get(a[i]));
            min.put(a[i],di);
            max.put(a[i],da);
        }
        boolean ok = false;
        for(Map.Entry<Integer,Integer> e :max.entrySet()){
            int k = e.getKey();
            int v = e.getValue();
            if(v>min.get(min.higherKey(k))){
                ok = true;
                break;
            }
        }
        if(ok) res = 0;
        else{
            HashMap<String,Integer> hm = new HashMap<String,Integer>();
            for(int i=0;i<n;i++){
                String s = String.valueOf(a[i])+"."+String.valueOf(b[i]);
                int d = 1;
                if(hm.containsKey(s)) d+=hm.get(s);
                hm.put(s,d);
            }
            for(int x:hm.values()){
                res *= fact[x];
                res %= mod;
            }
        }
        
        res = fact[n]+res;
        res %= mod;
        int[] num1 = new int[n+1];
        int[] num2 = new int[n+1];
        for(int i=0;i<n;i++){
            num1[a[i]]++;
            num2[b[i]]++;
        }
        long res1 = 1;
        long res2 = 1;
        for(int i=1;i<=n;i++){
            if(num1[i]>1){
                res1 *= fact[num1[i]];
                res1 %= mod;
            }
            if(num2[i]>1){
                res2 *= fact[num2[i]];
                res2 %= mod;
            }
        }
        res -= res1+res2;
        while(res<0) res += mod;
        out.println(res);
        out.flush();
   }
}