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.nio.CharBuffer;
import java.util.*;

public class P1207D {

    private static int MOD = 998244353;

    public static void main(String[] args) {
        SimpleScanner scanner = new SimpleScanner(System.in);
        PrintWriter writer = new PrintWriter(System.out);

        int n = scanner.nextInt();
        long[] frac = new long[n + 1];
        frac[0] = 1;
        for (int i = 1; i <= n; ++i)
            frac[i] = (frac[i - 1] * i) % MOD;
        ArrayList<int[]> x = new ArrayList<>(n);
        HashMap<Integer, Integer> aMap = new HashMap<>();
        HashMap<Integer, Integer> bMap = new HashMap<>();
        HashMap<Integer, int[]> aMinMax = new HashMap<>();
        HashMap<Integer, HashMap<Integer, Integer>> aBMap = new HashMap<>();
        for (int i = 0; i < n; ++i) {
            int a = scanner.nextInt();
            int b = scanner.nextInt();
            aMap.compute(a, (k, v) -> v == null ? 1 : v + 1);
            bMap.compute(b, (k, v) -> v == null ? 1 : v + 1);
            int[] minMax = aMinMax.get(a);
            if (minMax == null)
                aMinMax.put(a, new int[]{b, b});
            else {
                minMax[0] = Math.min(minMax[0], b);
                minMax[1] = Math.max(minMax[1], b);
            }
            HashMap<Integer, Integer> bInAMap = aBMap.get(a);
            if (bInAMap == null) {
                bInAMap = new HashMap<>();
                bInAMap.put(b, 1);
                aBMap.put(a, bInAMap);
            } else {
                bInAMap.compute(b, (k, v) -> v == null ? 1 : v + 1);
            }
        }
        long badA = 1;
        for (int v : aMap.values())
            badA = (badA * frac[v]) % MOD;
        long badB = 1;
        for (int v : bMap.values())
            badB = (badB * frac[v]) % MOD;
        ArrayList<int[]> minMaxList = new ArrayList<>(aMinMax.size());
        for (Map.Entry<Integer, int[]> entry : aMinMax.entrySet()) {
            minMaxList.add(new int[]{entry.getKey(), entry.getValue()[0], entry.getValue()[1]});
        }
        minMaxList.sort(Comparator.comparingInt(o -> o[0]));

        long duplicate = 0;
        boolean order = true;
        for (int i = 1; i < minMaxList.size(); ++i)
            if (minMaxList.get(i)[1] < minMaxList.get(i - 1)[2]) {
                order = false;
                break;
            }
        if (order) {
            duplicate = 1;
            for (HashMap<Integer, Integer> bInAMap : aBMap.values()) {
                long t = 1;
                for (int v : bInAMap.values())
                    duplicate = (duplicate * frac[v]) % MOD;
            }
        }
        writer.println((frac[n] - badA - badB + duplicate + 3L * MOD) % MOD);
        writer.close();
    }

    private static class SimpleScanner {

        private static final int BUFFER_SIZE = 10240;

        private Readable in;
        private CharBuffer buffer;
        private boolean eof;

        SimpleScanner(InputStream in) {
            this.in = new BufferedReader(new InputStreamReader(in));
            buffer = CharBuffer.allocate(BUFFER_SIZE);
            buffer.limit(0);
            eof = false;
        }


        private char read() {
            if (!buffer.hasRemaining()) {
                buffer.clear();
                int n;
                try {
                    n = in.read(buffer);
                } catch (IOException e) {
                    n = -1;
                }
                if (n <= 0) {
                    eof = true;
                    return '\0';
                }
                buffer.flip();
            }
            return buffer.get();
        }

        void checkEof() {
            if (eof)
                throw new NoSuchElementException();
        }

        char nextChar() {
            checkEof();
            char b = read();
            checkEof();
            return b;
        }

        String next() {
            char b;
            do {
                b = read();
                checkEof();
            } while (Character.isWhitespace(b));
            StringBuilder sb = new StringBuilder();
            do {
                sb.append(b);
                b = read();
            } while (!eof && !Character.isWhitespace(b));
            return sb.toString();
        }

        int nextInt() {
            return Integer.valueOf(next());
        }

        long nextLong() {
            return Long.valueOf(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}
