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.*;
public class Main {
    public static long mod = 998244353;
    public static long ans;
    public static class Pair {
        int a, b;
        public Pair(int a, int b) {
            this.a = a;
            this.b = b;
        }
    }
    public static long fac(int n) {
        long res = 1;
        for (long i = 2; i <= n; i++) {
            res *= i;
            res %= mod;
        }
        return res;
    }
    public static long perm(int n) {
        if (n == 1) return 1;
        return fac(n);
    }
    public static boolean check(Pair[] arr) {
        for (int i = 0; i < arr.length - 1; i++) {
            if (arr[i].a > arr[i + 1].a || arr[i].b > arr[i + 1].b) {
                return false;
            }
        }
        return true;
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        ans = fac(n);
        Pair[] arr = new Pair[n];
        for (int i = 0; i < n; i++) {
            Pair pair = new Pair(in.nextInt(), in.nextInt());
            arr[i] = pair;
        }
        Arrays.sort(arr, new Comparator<Pair>() {
            @Override
            public int compare(Pair a, Pair b) {
                if (a.a != b.a) {
                    return a.a - b.a;
                }
                return a.b - b.b;
            }
        });
        int elem = -1;
        int cnt = 1;
        long a, b, c;
        a = b = 1;
        c = 0;
        for (Pair pair : arr) {
            if (elem == -1) {
                elem = pair.a;
            } else if (pair.a != elem) {
                a *= perm(cnt);
                a %= mod;
                cnt = 1;
                elem = pair.a;
            } else {
                cnt++;
            }
        }
        a *= perm(cnt);
        a %= mod;
        Arrays.sort(arr, new Comparator<Pair>() {
            @Override
            public int compare(Pair a, Pair b) {
                if (a.b != b.b) {
                    return a.b - b.b;
                }
                return a.a - b.a;
            }
        });
        elem = -1;
        cnt = 1;
        for (Pair pair : arr) {
            if (elem == -1) {
                elem = pair.b;
            } else if (pair.b != elem) {
                b *= perm(cnt);
                b %= mod;
                cnt = 1;
                elem = pair.b;
            } else {
                cnt++;
            }
        }
        b *= perm(cnt);
        b %= mod;
        if (check(arr)) {
            c = 1;
            int ca = -1;
            int cb = -1;
            cnt = 1;
            for (Pair pair : arr) {
                if (ca == -1 && cb == -1) {
                    ca = pair.a;
                    cb = pair.b;
                } else if (pair.a != ca || pair.b != cb) {
                    c *= perm(cnt);
                    c %= mod;
                    ca = pair.a;
                    cb = pair.b;
                    cnt = 1;
                } else {
                    cnt++;
                }
            }
        }
        c *= perm(cnt);
        c %= mod;
        ans += 10 * mod;
        ans -= a;
        ans -= b;
        ans += c;
        ans %= mod;
        System.out.println(ans);
    }
}