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 Problem_1207D {
    final static long PRIME = 998244353;

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);

        int n = input.nextInt();
        List<Pair> list = new ArrayList<>();
        HashMap<Integer, Integer> firstMap = new HashMap<>();
        HashMap<Integer, Integer> secondMap = new HashMap<>();
        for (int i = 0; i < n; i++) {
            int a = input.nextInt(), b = input.nextInt();
            list.add(new Pair(a, b));
            firstMap.put(a, firstMap.get(a) == null ? 1 : firstMap.get(a) + 1);
            secondMap.put(b, secondMap.get(b) == null ? 1 : secondMap.get(b) + 1);
        }

        long numberOfPermutations = factorial(n);
        numberOfPermutations = (numberOfPermutations + PRIME - getNumberOfPermutations(firstMap.values())) % PRIME;
        numberOfPermutations = (numberOfPermutations + PRIME - getNumberOfPermutations(secondMap.values())) % PRIME;

        Collections.sort(list);

        boolean sorted = true;
        for (int i = 1; i < list.size(); i++) {
            if (list.get(i - 1).b > list.get(i).b) {
                sorted = false;
                break;
            }
        }
        if (sorted) {
            int count = 1;
            long p = 1;
            for (int i = 1; i < list.size(); i++) {
                if (list.get(i - 1).a == list.get(i).a && list.get(i - 1).b == list.get(i).b) {
                    count++;
                } else {
                    p = (p * factorial(count)) % PRIME;
                    count = 1;
                }
            }
            p = (p * factorial(count)) % PRIME;

            numberOfPermutations = (numberOfPermutations + PRIME + p) % PRIME;
        }

        System.out.println(numberOfPermutations);
        input.close();
    }

    static long getNumberOfPermutations(Collection<Integer> values) {
        long answer = 1;

        for (Integer value : values)
            answer = (answer * factorial(value)) % PRIME;

        return answer;
    }

    static long factorial(int n) {
        long fact = 1;
        for (int i = 2; i <= n; i++)
            fact = (fact * i) % PRIME;

        return fact;
    }

    static class Pair implements Comparable<Pair> {
        public long a, b;

        public Pair(long a, long b) {
            this.a = a;
            this.b = b;
        }

        public int compareTo(Pair p) {
            if (a > p.a)
                return 1;
            if (a < p.a)
                return -1;

            if (b > p.b)
                return 1;
            if (b < p.b)
                return -1;

            return 0;
        }
    }
}