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
{
    PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));

    long MOD = 998244353;

    public void go() throws IOException
    {
        long[] fact = new long[300005];
        long[] invFact = new long[300005];
        fact[0] = 1;
        invFact[0] = 1;
        for (int i = 1; i < fact.length; i++)
        {
            fact[i] = i * fact[i-1] % MOD;
            invFact[i] = inverse(i) * invFact[i-1] % MOD;
        }

        StringTokenizer tok = new StringTokenizer(in.readLine());
        int n = Integer.parseInt(tok.nextToken());
        ArrayList<int[]> pairs = new ArrayList<>();
        for (int i = 0; i < n; i++)
        {
            tok = new StringTokenizer(in.readLine());
            pairs.add(new int[] {Integer.parseInt(tok.nextToken()), Integer.parseInt(tok.nextToken())});
        }
        Collections.sort(pairs, (one, two) -> {
            if (one[0] == two[0])
            {
                return Integer.compare(one[1], two[1]);
            }
            return Integer.compare(one[0], two[0]);
        });
        long inv = 1;
        int[] prev = pairs.get(0);
        int[] count = {1, 1};
        for (int i = 1; i < n; i++)
        {
            int[] curr = pairs.get(i);
            if (prev[0] != curr[0])
            {
                if (count[0] > 1)
                {
                    inv *= fact[count[0]];
                    inv %= MOD;
                }
                count[0] = 0;
            }
            count[0]++;
            prev = curr;
        }
        if (count[0] > 1)
        {
            inv *= fact[count[0]];
            inv %= MOD;
        }
        count[0] = 1;
        ArrayList<int[]> pairs2 = new ArrayList<>(pairs);
        Collections.sort(pairs, (one, two) -> {
            if (one[1] == two[1])
            {
                return Integer.compare(one[0], two[0]);
            }
            return Integer.compare(one[1], two[1]);
        });

        long inv2 = 1;
        prev = pairs.get(0);
        for (int i = 1; i < n; i++)
        {
            int[] curr = pairs.get(i);
            if (prev[1] != curr[1])
            {
                if (count[1] > 1)
                {
                    inv2 *= fact[count[1]];
                    inv2 %= MOD;
                }
                count[1] = 0;
            }
            count[1]++;
            prev = curr;
        }
        if (count[1] > 1)
        {
            inv2 *= fact[count[1]];
            inv2 %= MOD;
        }
        count[1] = 1;

        
        long dupInv = 1;
        for (int i = 0; i < n; i++)
        {
            if (pairs.get(i)[0] != pairs2.get(i)[0] || pairs.get(i)[1] != pairs2.get(i)[1])
            {
                dupInv--;
                break;
            }
        }
        prev = pairs.get(0);
        for (int i = 1; i < n; i++)
        {
            int[] curr = pairs.get(i);
            if (curr[0] != prev[0] || curr[1] != prev[1])
            {
                dupInv *= fact[count[0]];
                dupInv %= MOD;
                count[0] = 0;
            }
            count[0]++;
            prev = curr;
        }
        dupInv *= fact[count[0]];
        dupInv %= MOD;
        count[0] = 1;

        // out.println(inv + inv2);
        // out.println(dupInv);
        long ans = fact[n];
        ans = (ans - inv + MOD) % MOD;
        ans = (ans - inv2 + MOD) % MOD;
        ans = (ans + dupInv + MOD) % MOD;
        out.println(ans);

        out.flush();
        in.close();
    }

    public long inverse(long n)
    {
        return modPow(n, MOD-2);
    }

    public long modPow(long n, long p)
    {
        if (p == 0)
        {
            return 1;
        }
        long ans = modPow(n*n%MOD, p/2);
        if (p % 2 == 1)
        {
            ans *= n;
            ans %= MOD;
        }
        return ans;
    }

    public static void main(String[] args) throws IOException
    {
        new D().go();
    }
}
