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.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;

public class CF1207D {

    private static int MOD = 998244353;

    //good数 = 排列总数 - 左节点升序排列的总数 - 右节点升序排列的总数 + 左右节点同时升序排列的总数
    //排列总数 = 数组长度的阶乘
    //升序排列的总数 = 所有出现重复情况数量的阶乘的乘积

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[][] a = new int[n][2];
        for(int i = 0; i < n; i++){
            a[i][0] = sc.nextInt();
            a[i][1] = sc.nextInt();
        }
        long allCnt = getAllCnt(a);
        long leftBadCnt = getLeftBadCnt(a);
        long rightBadCnt = getRightBadCnt(a);
        long bothBadCnt = getBothBadCnt(a);
        System.out.println((allCnt - leftBadCnt - rightBadCnt + bothBadCnt + MOD + MOD) % MOD);
    }

    public static long getAllCnt(int[][] a){
        long r = 1;
        for(int i = 1; i <= a.length; i++){
            r = r * i % MOD;
        }
        return r;
    }

    public static long getLeftBadCnt(int[][] a){
        Arrays.sort(a, new Comparator<int[]>() {
            public int compare(int[] one, int[] two) {
                if(one[0] < two[0]){
                    return -1;
                }else if(one[0] == two[0]){
                    return 0;
                }else{
                    return 1;
                }
            }
        });

        long r = 1;
        int k = 1;
        for(int i = 1; i < a.length; i++){
            if(a[i][0] == a[i-1][0]){
                k++;
                r = r * k % MOD;
            }else{
                k = 1;
            }
        }
        return r;
    }

    public static long getRightBadCnt(int[][] a){
        Arrays.sort(a, new Comparator<int[]>() {
            public int compare(int[] one, int[] two) {
                if(one[1] < two[1]){
                    return -1;
                }else if(one[1] == two[1]){
                    return 0;
                }else{
                    return 1;
                }
            }
        });

        long r = 1;
        int k = 1;
        for(int i = 1; i < a.length; i++){
            if(a[i][1] == a[i-1][1]){
                k++;
                r = r * k % MOD;
            }else{
                k = 1;
            }
        }
        return r;
    }

    public static long getBothBadCnt(int[][] a){
        Arrays.sort(a, new Comparator<int[]>() {
            public int compare(int[] one, int[] two) {
                if(one[0] < two[0]){
                    return -1;
                }else if(one[0] > two[0]){
                    return 1;
                }else{
                    if(one[1] < two[1]){
                        return -1;
                    }else if(one[1] == two[1]){
                        return 0;
                    }else{
                        return 1;
                    }
                }
            }
        });

        long r = 1;
        int k = 1;
        for(int i = 1; i < a.length; i++){
            if(a[i][1] < a[i-1][1]){
                r = 0;
                break;
            }else if(a[i][0] == a[i-1][0] && a[i][1] == a[i-1][1]){
                k++;
                r = r * k % MOD;
            }else{
                k = 1;
            }
        }

        return r;
    }

}
