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 Main {
 
    public static long MOD = 998_244_353;
    public static void main(String[] args) {
 
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        Pair[] s = new Pair[n];
        Map<Integer, Integer> mapA = new HashMap();
        Map<Integer, Integer> mapB = new HashMap();

        for(int i = 0; i < n; ++i){
            int a = scanner.nextInt();
            int b = scanner.nextInt();
            s[i] = new Pair(a, b);
            
            mapA.put(a, mapA.getOrDefault(a, 0) + 1);
            mapB.put(b, mapB.getOrDefault(b, 0) + 1);

        }

        long[] fact = new long[n + 1];
        fact[0] = 1;
        for (int i = 1; i <= n ; ++i) {
            fact[i] = (i * fact[i - 1]) % MOD;
        }

        long A = 1;
        for ( Map.Entry<Integer, Integer> e : mapA.entrySet()) {
            A = (A * fact[e.getValue()]) % MOD;
        }

        long B = 1;
        for ( Map.Entry<Integer, Integer> e : mapB.entrySet()) {
            B = (B * fact[e.getValue()]) % MOD;
        }

        Arrays.sort(s, (p1, p2) -> {
            if (p1.a != p2.a) {
                return p1.a - p2.a;
            }
            return p1.b - p2.b;
       });  

        long C = 1;
        int l = 0;
        while(l < n){
            int r = l + 1;
            while (r < n && s[r].a == s[l].a) ++r;
            Map<Integer, Integer> m = new HashMap();
            for(int i = l; i < r; ++i) m.put(s[i].b, m.getOrDefault(s[i].b, 0) + 1);
            for (int k : m.values()) {
                C = (C * fact[k]) % MOD;
            }
            l = r;
        }

        for(int i = 1; i < n; ++i){
            if (s[i].b < s[i - 1].b) {
                C = 0;
            }
        }
  
        long ans = (fact[n] - A - B) % MOD;
        ans = (ans + C) % MOD;
        if (ans < 0) {
            ans += MOD;
        }
        System.out.println(ans);

    }

    static class Pair{
        int a;
        int b;
        public Pair(int a, int b){
            this.a = a;
            this.b = b;
        }
    }
}