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.PrintWriter;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class NumberOfPermutations {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int t = 1;
        for (int i = 0; i < t; i++) {
            solve(sc, pw);
        }
        pw.close();
    }
    static int mod = 998244353;
    static void solve(Scanner in, PrintWriter out){
        int n = in.nextInt();
        int[][] arr = new int[n][2];
        for (int i = 0; i < n; i++) {
            arr[i] = new int[]{in.nextInt(), in.nextInt()};
        }
//        long ans = 0;
        Map<Integer, Integer> first = new HashMap<>();
        Map<Integer, Integer> second = new HashMap<>();
        for (int i = 0; i < n; i++) {
            first.put(arr[i][0], first.getOrDefault(arr[i][0], 0) + 1);
            second.put(arr[i][1], second.getOrDefault(arr[i][1], 0) + 1);
        }
        long[] pre = new long[n + 1];
        pre[0] = 1;
        for (int i = 1; i <= n; i++) {
            pre[i] = (pre[i - 1] * i) % mod;
        }
        long tot = 1;
        for (int i = n; i >= 1; i--) {
            tot = (tot * i) % mod;
        }
        long fs = 1, sc = 1, cb = 1;
        for (int i = 1; i <= n; i++) {
            if (first.containsKey(i)){
                fs = (fs * pre[first.get(i)]) % mod;
            }
            if (second.containsKey(i)){
                sc = (sc * pre[second.get(i)]) % mod;
            }
        }
        Arrays.sort(arr, (a, b) -> {
            if (a[0] == b[0]) return a[1] - b[1];
            return a[0] - b[0];
        });
        boolean find = true;
        for (int i = 0; i < n - 1; i++) {
            if (arr[i][1] > arr[i + 1][1]){
                find = false;
                break;
            }
        }
        if (find){
            for (int i = 0; i < n; i++) {
                int j = i;
                while (j < n && arr[j][0] == arr[i][0] && arr[j][1] == arr[i][1]) j++;
                int cnt = j - i;
                cb = (cb * pre[cnt]) % mod;
                i = j - 1;
            }
        }else{
            cb = 0;
        }
        tot = (((tot - fs - sc + cb) % mod) + mod) % mod;
        out.println(tot);
    }
}
