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.Arrays;
import java.util.Comparator;
import java.util.Scanner;

public class D71 {
    static long mod = 998244353;
    static long fact[] = new long[(int) 5e5];

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        pair[] t = new pair[n];
        for (int i = 0; i < n; i++) {
            t[i] = new pair(in.nextInt(), in.nextInt());
        }
        long u = 1;
        long v = 1;
        long w = 1;
        fact();
        Arrays.sort(t, new Comparator<pair>() {
            @Override
            public int compare(pair p1, pair p2) {
                if (p1.a != p2.a) {
                    return p1.a - p2.a;
                }
                return p1.b - p2.b;
            }
        });
        int cur = 1;
        for (int i = 1; i < n; i++) {
            if (t[i].a == t[i - 1].a) {
                cur++;
            } else {
                u = (u * fact[cur]) % mod;
                cur = 1;
            }
        }
        u = (u * fact[cur]) % mod;
        cur = 1;
        Arrays.sort(t, new Comparator<pair>() {
            @Override
            public int compare(pair p1, pair p2) {
                if (p1.b != p2.b) {
                    return p1.b - p2.b;
                }
                return p1.a - p2.a;
            }
        });
        for (int i = 1; i < n; i++) {
            if (t[i].b == t[i - 1].b) {
                cur++;
            } else {
                v = (v * fact[cur]) % mod;
                cur = 1;
            }
        }
        v = (v * fact[cur]) % mod;
        cur = 1;
        for (int i = 1; i < n; i++) {
            if (t[i].a < t[i - 1].a) {
                w = 0;
            }
            if (t[i].equals(t[i - 1])) {
                cur++;
            } else {
                w = (w * fact[cur]) % mod;
                cur = 1;
            }
        }
        w = (w * fact[cur]) % mod;
        long ans = ((fact[n] - u - v + w) % mod + mod) % mod;
        System.out.println(ans);
    }

    static void fact() {
        fact[0] = 1;
        for (int i = 1; i < fact.length; i++) {
            fact[i] = ((fact[i - 1] * i) % mod + mod) % mod;
        }
    }

    static long div(long x) {
        return pow(x, mod - 2);
    }

    static long pow(long x, long m) {
        if (m == 0) {
            return 1;
        }
        long buf = pow(x, m / 2);
        buf = (buf * buf) % mod;
        if (m % 2 == 1) {
            buf = (x * buf) % mod;
        }
        return buf;
    }

    static class pair {
        int a;
        int b;

        pair(int a, int b) {
            this.a = a;
            this.b = b;
        }

        boolean equals(pair p) {
            return a == p.a && b == p.b;
        }
    }
}
