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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Vector;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.util.Collections;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author xwchen
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskD {
        long mod = 998244353L;

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            Vector<Pair> pairs = new Vector<>();
            int[] a = new int[n + 1];
            int[] b = new int[n + 1];
            long[] f = new long[n + 1];
            long c3 = 1;
            f[0] = 1;
            for (int i = 0; i < n; ++i) {
                f[i + 1] = f[i] * (i + 1) % mod;
                int x = in.nextInt();
                int y = in.nextInt();
                a[x]++;
                b[y]++;
                pairs.add(new Pair(x, y));
            }

            Collections.sort(pairs);
            long res = f[n];
            long c1 = 1;
            long c2 = 1;
            for (int i = 1; i <= n; ++i) {
                c1 = c1 * f[a[i]] % mod;
                c2 = c2 * f[b[i]] % mod;
                //System.out.println(pairs.get(i-1).x+" "+pairs.get(i-1).y);
            }
            int last = 1;
            boolean OK = true;
            for (int i = 1; i < n; ++i) {
                if (pairs.get(i - 1).y > pairs.get(i).y) {
                    OK = false;
                }
                if (pairs.get(i - 1).x == pairs.get(i).x && pairs.get(i - 1).y == pairs.get(i).y) {
                    last++;
                } else {
                    c3 = c3 * f[last] % mod;
                    last = 1;
                }
            }
            c3 = c3 * f[last] % mod;
            if (!OK) c3 = 0;
            //System.out.println(c1 + " " + c2 + " " + c3);
            res = ((res - c1 - c2 + c3) % mod + mod) % mod;
            out.println(res);

        }

        public class Pair implements Comparable<Pair> {
            int x;
            int y;

            public Pair(int x, int y) {
                this.x = x;
                this.y = y;
            }

            public int compareTo(Pair o) {
                if (o.x == x) return y - o.y;
                return x - o.x;
            }

        }

    }

    static class InputReader {
        private BufferedReader reader;
        private StringTokenizer tokenizer = new StringTokenizer("");

        public InputReader(InputStream inputStream) {
            this.reader = new BufferedReader(
                    new InputStreamReader(inputStream));
        }

        public String next() {
            while (!tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}

