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. 
/**
 * @author derrick20
 */
import java.io.*;
import java.util.*;

public class NumberPermutations {
    public static void main(String args[]) throws Exception {
        FastScanner sc = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int N = sc.nextInt();
        HashMap<Integer, Integer> freqA = new HashMap<>();
        HashMap<Integer, Integer> freqB = new HashMap<>();

        Pair[] arr = new Pair[N];
        for (int i = 0; i < arr.length; i++) {
            int a = sc.nextInt(); int b = sc.nextInt();
            update(freqA, a);
            update(freqB, b);
            arr[i] = new Pair(a, b);
        }

        long mod = 998244353;
        long[] factorial = new long[N + 1];
        factorial[0] = 1;
        for (int i = 1; i <= N; i++) {
            factorial[i] = (factorial[i - 1] * (long) i) % mod;
        }

        long aWays = 1;
        long bWays = 1;
        for (int val : freqA.keySet()) {
            int freq = freqA.get(val);
            aWays *= factorial[freq];
            aWays %= mod;
        }

        for (int val : freqB.keySet()) {
            int freq = freqB.get(val);
            bWays *= factorial[freq];
            bWays %= mod;
        }
//        System.out.println(freqA);
//        System.out.println(freqB);
        Arrays.sort(arr);
        boolean bothSorted = true;
        HashMap<String, Integer> freqBoth = new HashMap<>();
//        System.out.println(Arrays.toString(arr));
        for (int i = 1; i < N; i++) {
            if (!(arr[i - 1].a <= arr[i].a && arr[i - 1].b <= arr[i].b)) {
                bothSorted = false;
                break;
                // not all adjacents are increasing
            }
        }
        long bothWays = 0;
        if (bothSorted) {
            for (Pair p : arr) {
                String key = p.toString();
                if (freqBoth.containsKey(key)) {
                    freqBoth.put(key, freqBoth.get(key) + 1);
                }
                else {
                    freqBoth.put(key, 1);
                }
            }
            bothWays = 1;
            for (String val : freqBoth.keySet()) {
                int freq = freqBoth.get(val);
                bothWays *= factorial[freq];
                bothWays %= mod;
            }
        }
//        System.out.printf("A = %d B = %d Both = %d \n", aWays, bWays, bothWays);
        long bad = (aWays + bWays - bothWays + mod) % mod;
        long good = (factorial[N] - bad + mod) % mod;
        out.println(good);

        out.close();
    }

    static void update2(HashMap<String, Integer> map, String val) {
        if (map.containsKey(val)) {
            map.put(val, map.get(val) + 1);
        }
        else {
            map.put(val, 1);
        }
    }

    static void update(HashMap<Integer, Integer> map, int val) {
        if (map.containsKey(val)) {
            map.put(val, map.get(val) + 1);
        }
        else {
            map.put(val, 1);
        }
    }

    static class Pair implements Comparable<Pair> {
        int a, b;
        public Pair(int a, int b) {
            this.a = a; this.b = b;
        }

        public String toString() {
            return a + "#" + b;
        }

        public int compareTo(Pair p2) {
            if (a != p2.a) {
                return a - p2.a;
            }
            else {
                return b - p2.b;
            }
        }
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        FastScanner(FileReader s) {
            br = new BufferedReader(s);
        }

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

        String nextLine() throws IOException { return br.readLine(); }

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

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

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