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.*;
import java.util.*;

public class Main {

    public static void main(String[] args) throws Exception {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        int n = Integer.parseInt(in.readLine());
        ArrayList<Pair> list = new ArrayList<>();
        HashMap<Integer, Integer> left = new HashMap<>();
        HashMap<Integer, Integer> right = new HashMap<>();
        HashMap<Pair, Integer> both = new HashMap<>();
        for (int i = 0; i < n; i++) {
            StringTokenizer st = new StringTokenizer(in.readLine());
            int x = Integer.parseInt(st.nextToken());
            int y = Integer.parseInt(st.nextToken());
            list.add(new Pair(x, y));
            if (!left.containsKey(x)) left.put(x, 0);
            if (!right.containsKey(y)) right.put(y, 0);
            if (!both.containsKey(new Pair(x,y))) both.put(new Pair(x,y), 0);
            left.put(x, left.get(x)+1);
            right.put(y, right.get(y)+1);
            both.put(new Pair(x,y), both.get(new Pair(x,y))+1);
        }
        Collections.sort(list);

        long[] fac = new long[300001];
        long mod = 998244353;
        fac[0] = 1;
        for (int i = 1; i < 300001; i++) {
            fac[i] = fac[i-1]*i;
            fac[i] %= mod;
           // System.out.println(fac[i]);
        }

        long leftPerm = 1;
        for (int key : left.keySet()) {
            leftPerm *= fac[left.get(key)];
            leftPerm %= mod;
        }
        long rightPerm = 1;
        for (int key : right.keySet()) {
            rightPerm *= fac[right.get(key)];
            rightPerm %= mod;
        }

        long bothPerm = 1;
        for (Pair key : both.keySet()) {
            bothPerm *= fac[both.get(key)];
            bothPerm %= mod;
        }

        boolean flag = false;
        for (int i = 0; i < n-1; i++) {
            if (list.get(i).x > list.get(i+1).x || list.get(i).y > list.get(i+1).y) {
                flag = true;
            }
        }
        if (flag) {
            bothPerm = 0;
        }

        //System.out.println("LEFT: " + leftPerm + " RIGHT: " + rightPerm + " BOTH: " + bothPerm);
        long total = leftPerm+rightPerm-bothPerm;
        while (total < 0) total += mod;
        long res = fac[n] - total;
        while (res < 0) res += mod;
        res %= mod;
        out.println(res);
        out.close();
    }

    static class Pair implements Comparable<Pair>{
        int x, y;
        public Pair(int x, int y) {
            this.x = x;
            this.y = y;
        }
        @Override
        public int compareTo(Pair p) {
            if (this.x == p.x) {
                return this.y-p.y;
            } else {
                return this.x-p.x;
            }
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Pair pair = (Pair) o;
            return x == pair.x &&
                    y == pair.y;
        }

        @Override
        public int hashCode() {
            return Objects.hash(x, y);
        }
    }
}

/*

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