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 D_NumberOfPermutations {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader inp = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Solver solver = new Solver();
        solver.solve(inp, out);
        out.close();
    }

    private static class Solver {
        private void solve(InputReader inp, PrintWriter out) {
            int n = inp.nextInt();
            int[] cntA = new int[n+1];
            int[] cntB = new int[n+1];
            HashMap<Pair, Integer> cnt = new HashMap<>();
            ArrayList<Pair> pairs = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                int a = inp.nextInt(), b = inp.nextInt();
                Pair p = new Pair(a, b);
                cntA[a]++;
                cntB[b]++;
                cnt.put(p, cnt.getOrDefault(p, 0) + 1);
                pairs.add(p);
            }
            Collections.sort(pairs);
            boolean sorted = true;
            for (int i = 0; i < n - 1; i++) {
                Pair p1 = pairs.get(i);
                Pair p2 = pairs.get(i+1);
                if (p1.a > p2.a || p1.b > p2.b) sorted = false;
            }

            long[] fac = new long[n+1];
            long MOD = 998244353;
            fac[0] = 1;
            for (int i = 1; i <= n; i++) fac[i] = (fac[i-1] * i) % MOD;

            long badA = 1;
            for (int i = 1; i <= n; i++) badA = (badA * fac[cntA[i]]) % MOD;
            //out.println("bad A: " + badA);
            long badB = 1;
            for (int i = 1; i <= n; i++) badB = (badB * fac[cntB[i]]) % MOD;
            //out.println("bad B: " + badB);

            long bad = (badA + badB) % MOD;
            //out.println("sorted: " + sorted);
            if (sorted) {
                long sub = 1;
                for (Pair p: cnt.keySet()) {
                    sub = (sub * fac[cnt.get(p)]) % MOD;
                }
                //out.println("duplicate: " + sub);
                bad -= sub;
            }


            long res = (fac[n] - bad) % MOD;
            out.print((res + MOD) % MOD);
        }

        public class Pair implements Comparable<Pair> {
            int a, b;
            int hash;

            public Pair(int a, int b) {
                this.a = a;
                this.b = b;
                hash = 527 + a;
                hash = hash * 31 + b;
            }

            @Override
            public int hashCode() {
                return hash;
            }

            @Override
            public boolean equals(Object o) {
                if (!(o instanceof Pair)) return false;
                Pair o2 = (Pair) o;
                return o2.a == this.a && o2.b == this.b;
            }

            @Override
            public int compareTo(Pair p) {
                if (a != p.a) return Integer.compare(a, p.a);
                return Integer.compare(b, p.b);
            }
        }
    }

    static class InputReader {
        BufferedReader reader;
        StringTokenizer tokenizer;

        InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
            return Long.parseLong(next());
        }
    }
}