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 Solution {

	static class pair implements Comparable<pair> {
		int a, b;

		pair(int x, int y) {
			a = x;
			b = y;
		}

		@Override
		public int compareTo(pair y) {
			// TODO Auto-generated method stub
			return a != y.a ? a - y.a : b - y.b;
		}
	}

	static int MOD = 998244353;

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner();
		PrintWriter out = new PrintWriter(System.out);
		int n = sc.nextInt();
		pair[] pairs = new pair[n];
		int[] cnt = new int[n];
		long[] fac = new long[n + 5];
		fac[0] = 1;
		for (int i = 1; i < fac.length; i++)
			fac[i] = fac[i - 1] * i % MOD;
		TreeMap<pair, Integer> map = new TreeMap();
		for (int i = 0; i < n; i++) {
			pairs[i] = new pair(sc.nextInt() - 1, sc.nextInt() - 1);
			cnt[pairs[i].a]++;
			map.put(pairs[i], map.getOrDefault(pairs[i], 0) + 1);

		}
		long ans = fac[n];
		long sortedByFirst = 1;
		for (int i = 0; i < n; i++) {
			sortedByFirst = sortedByFirst * fac[cnt[i]] % MOD;
		}
		ans = sub(ans, sortedByFirst);
		long sortedBySec = 1;
		Arrays.fill(cnt, 0);
		for (pair p : pairs)
			cnt[p.b]++;
		for (int i = 0; i < n; i++) {
			sortedBySec = sortedBySec * fac[cnt[i]] % MOD;
		}
		ans = sub(ans, sortedBySec);
		Arrays.fill(cnt, 0);
		boolean sortedFirst = true, sortedSecond = true;
		for (pair p : pairs)
			if (p.a == p.b) {
				cnt[p.a]++;
			}
		Arrays.sort(pairs, (x, y) -> x.a != y.a ? x.a - y.a : x.b - y.b);
		for (int i = 0; i + 1 < n; i++) {
			if (pairs[i + 1].b < pairs[i].b)
				sortedSecond = false;
		}
		if (sortedFirst && sortedSecond) {
			long both = 1;
			for (int x : map.values())
				both = both * fac[x] % MOD;
			ans += both;
			ans %= MOD;
		}
		out.println(ans);
		out.close();

	}

	static long sub(long x, long y) {
		x -= y;
		if (x < 0)
			x += MOD;
		return x;
	}

	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

		Scanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		Scanner(String fileName) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(fileName));
		}

		String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		String nextLine() throws IOException {
			return br.readLine();
		}

		int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		long nextLong() throws NumberFormatException, IOException {
			return Long.parseLong(next());
		}

		double nextDouble() throws NumberFormatException, IOException {
			return Double.parseDouble(next());
		}

		boolean ready() throws IOException {
			return br.ready();
		}

	}

	static void sort(int[] a) {
		shuffle(a);
		Arrays.sort(a);
	}

	static void shuffle(int[] a) {
		int n = a.length;
		Random rand = new Random();
		for (int i = 0; i < n; i++) {
			int tmpIdx = rand.nextInt(n);
			int tmp = a[i];
			a[i] = a[tmpIdx];
			a[tmpIdx] = tmp;
		}
	}

}