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 {
    static int inf = (int) 1e9 + 7;
    public static void main(String[] args) throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(System.out);
        int n = nextInt();
        int mod =   998244353;
        long fact[] = new long [n + 1];
        fact[0] = 1;
        for(int i = 1;i <= n;i++) fact[i] = (fact[i - 1] * i) % mod;
        long ans = 0;
        pair a[] = new pair[n];
        for(int i = 0;i < n;i++) a[i] = new pair(nextInt(), nextInt());
        Arrays.sort(a, new pair());
        boolean bad = false;
        int last = 0;
        for(int i = 0;i < n;i++) {
            if (last > a[i].b) {
                bad = true;
            }
            last = a[i].b;
        }
        if (!bad) {
            int cnt = 1;
            long u = 1;
            for(int i = 1;i < n;i++) {
                if (a[i].a != a[i - 1].a || a[i].b != a[i - 1].b) {
                    u = (u * fact[cnt]) % mod;
                    cnt = 1;
                }else cnt++;
            }
            u = (u * fact[cnt]) % mod;
            ans += u;
        }
        long u = 1;
        int cnt = 1;
        for(int i = 1;i < n;i++) {
            if (a[i].a != a[i - 1].a) {
                u = (u * fact[cnt]) % mod;
                cnt = 1;
            }else cnt++;
        }
        u = (u * fact[cnt]) % mod;
        ans -= u;

        cnt = 1;
        u = 1;

        for(int i = 0;i < n;i++) {
            int t = a[i].a;
            a[i].a = a[i].b;
            a[i].b = t;
        }
        Arrays.sort(a, new pair());

        for(int i = 1;i < n;i++) {
            if (a[i].a != a[i - 1].a) {
                u = (u * fact[cnt]) % mod;
                cnt = 1;
            }else cnt++;
        }
        u = (u * fact[cnt]) % mod;
        ans -= u;
        ans += fact[n];
        ans = (ans % mod + mod) % mod;
        pw.println(ans);
        pw.close();
    }

    static BufferedReader br;
    static StringTokenizer st = new StringTokenizer("");
    static PrintWriter pw;

    static String next() throws IOException {
        while (!st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
        return st.nextToken();
    }

    static int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    static long nextLong() throws IOException {
        return Long.parseLong(next());
    }
}

class pair implements Comparator<pair>{
    int a, b;
    pair(int x, int y) {
        a = x;
        b = y;
    }

    pair() {}

    @Override
    public int compare(pair pair, pair t1) {
        return Integer.compare(pair.a, t1.a) * 2 + Integer.compare(pair.b, t1.b);
    }
}