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.*;

import static java.lang.Math.*;

public class D {
    public static void main(String[] args) throws Exception {
        Scanner in = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        long ans = 0, mod = 998244353;
        int n = in.nextInt();
        Pair[] pairs = new Pair[n];
        long fac[] =  new long[n + 1];
        fac[0] = 1;
        for (int i = 0; i < n; i++) {
            pairs[i] = new Pair(in.nextInt(), in.nextInt());
            fac[i + 1] = (fac[i] * (i + 1)) % mod;
        }
        Arrays.sort(pairs);

        long same = 1;
        for (int i = 1; i < n; i++) {
            same &= (pairs[i].f >= pairs[i - 1].f && pairs[i].s >= pairs[i - 1].s) ? 1 : 0;
        }
        for (int i = 0, c = 0; i < n; i++) {
            if (i == 0 || pairs[i].compareTo(pairs[i - 1]) == 0) {
                c++;
            } else {
                same = (same * fac[c]) % mod;
                c = 1;
            }
            if (i > 0 && i == n - 1) {
                same = (same * fac[c]) % mod;
                c = 1;
            }
        }

        Arrays.sort(pairs, new Comparator<Pair>() {
            public int compare(Pair o1, Pair o2) {
                return Integer.compare(o1.f, o2.f);
            }
        });

        long a = 1, b = 1;
        for (int i = 0, c = 0; i < n; i++) {
            if (i == 0 || pairs[i].f == pairs[i - 1].f) {
                c++;
            } else {
                a = (a * fac[c]) % mod;
                c = 1;
            }
            if (i > 0 && i == n - 1) {
                a = (a * fac[c]) % mod;
                c = 1;
            }
        }

        Arrays.sort(pairs, new Comparator<Pair>() {
            public int compare(Pair o1, Pair o2) {
                return Integer.compare(o1.s, o2.s);
            }
        });

        for (int i = 0, c = 0; i < n; i++) {
            if (i == 0 || pairs[i].s == pairs[i - 1].s) {
                c++;
            } else {
                b = (b * fac[c]) % mod;
                c = 1;
            }
            if (i > 0 && i == n - 1) {
                b = (b * fac[c]) % mod;
                c = 1;
            }
        }

        ans = (fac[n] - a - b + same) % mod;
        ans = (ans + mod) % mod;
        pw.println(ans);

        pw.close();
    }

    static void debug(Object... obj) {
        System.err.println(Arrays.deepToString(obj));
    }

    static class Pair implements Comparable<Pair> {
        int f, s;

        Pair(int F, int S) {
            f = F;
            s = S;
        }

        public int compareTo(Pair p) {
            if (f == p.f) return s - p.s;
            return f - p.f;
        }

        public String toString() {
            return f + " " + s;
        }
    }
}