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.*;
import java.io.*;

public class D {
    static final long mod = 998244353;
    public static void main(String[] args) {
        FastScanner scanner = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int n = scanner.nextInt();
        PairA[] a = new PairA[n];
        PairB[] b = new PairB[n];
        for(int i = 0; i < n; i++) {
            a[i] = new PairA(scanner.nextInt(), scanner.nextInt(), i);
            b[i] = new PairB(a[i].a, a[i].b, a[i].c);
        }
        Arrays.sort(a);
        Arrays.sort(b);
        long[] f = new long[n + 1];
        f[0] = 1;
        for(int i = 1; i <= n; i++) f[i] = (f[i-1] * i ) % mod;
        long ans1 = 1;
        long ans2 = 1;
        int ac = 1;
        int bc = 1;
        for(int i = 1; i < n; i++) {
            if (a[i].a == a[i-1].a) ac++;
            else {
                ans1 = (ans1 * f[ac]) % mod;
                ac = 1;
            }
            if (b[i].b == b[i-1].b) bc++;
            else {
                ans2 = (ans2 * f[bc]) % mod;
                bc = 1;
            }
        }
        ans2 = (ans2 * f[bc]) % mod;
        ans1 = (ans1 * f[ac]) % mod;
        long tot = (ans1 + ans2) % mod;
        int cc = 0;
        for(int i = 0; i < n; i++) {
            if (a[i].a == b[i].a && a[i].b == b[i].b) cc++;
        }
        if (cc == n) {
            long ans = 1;
            int c = 1;
            for(int i = 1; i < n; i++) {
                if (b[i].b == b[i-1].b && a[i].a == a[i-1].a) c++;
                else {
                    ans = (ans * f[c]) % mod;
                    c = 1;
                }
            }
            ans = (ans * f[c]) % mod;
            tot = (tot + mod - ans) % mod;
        }
        out.println((f[n] + mod - tot) % mod);
        out.flush();
    }
    
    static class PairA implements Comparable<PairA> {
        int a,b,c;
        public PairA(int aa, int bb, int cc) {
            a = aa; b = bb; c = cc;
        }
        public int compareTo(PairA o) {
            if (a == o.a) {
                if (b == o.b) {
                    return c - o.c;
                }
                return b - o.b;
            }
            return a - o.a;
        }
    }
    static class PairB implements Comparable<PairB> {
        int a,b,c;
        public PairB(int aa, int bb, int cc) {
            a = aa; b = bb; c = cc;
        }
        public int compareTo(PairB o) {
            if (b == o.b) {
                if (a == o.a) {
                    return c - o.c;
                }
                return a - o.a;
            }
            return b - o.b;
        }
    }

    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
        
        public FastScanner(Reader in) {
            br = new BufferedReader(in);
        }
        
        public FastScanner() {
            this(new InputStreamReader(System.in));
        }
        
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        int nextInt() {
            return Integer.parseInt(next());
        }
        
        long nextLong() {
            return Long.parseLong(next());
        }
        
        double nextDouble() {
            return Double.parseDouble(next());
        }
        
        String readNextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}
