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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.util.Comparator;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Washoum
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        inputClass in = new inputClass(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        DNumberOfPermutations solver = new DNumberOfPermutations();
        solver.solve(1, in, out);
        out.close();
    }

    static class DNumberOfPermutations {
        static final int mod = 998244353;

        public void solve(int testNumber, inputClass sc, PrintWriter out) {
            int n = sc.nextInt();
            if (n == 1) {
                out.println(0);
                return;
            }
            DNumberOfPermutations.Pair[] tab = new DNumberOfPermutations.Pair[n];
            for (int i = 0; i < n; i++) {
                tab[i] = new DNumberOfPermutations.Pair(sc.nextInt(), sc.nextInt());
            }
            long[] fact = new long[n + 1];
            fact[0] = 1;
            for (int i = 1; i <= n; i++) {
                fact[i] = fact[i - 1] * i;
                fact[i] %= mod;
            }
            long ans = fact[n];
            Arrays.sort(tab, Comparator.comparingInt(a -> a.x));
            long tosub = 1;
            for (int i = 0; i < n; i++) {
                int act = i;
                while (act < n && tab[act].x == tab[i].x) act++;
                act--;
                if (act - i > 0) {

                    tosub *= fact[act - i + 1];
                    tosub %= mod;

                }
                i = act;
            }
            ans -= tosub;
            if (ans < 0)
                ans += mod;
            Arrays.sort(tab, Comparator.comparingInt(a -> a.y));
            tosub = 1;
            for (int i = 0; i < n; i++) {
                int act = i;
                while (act < n && tab[act].y == tab[i].y) act++;
                act--;
                if (act - i > 0) {
                    tosub *= fact[act - i + 1];
                    tosub %= mod;

                }
                i = act;
            }
            ans -= tosub;
            if (ans < 0)
                ans += mod;
            Arrays.sort(tab, (a, b) -> {
                if (a.x == b.x) {
                    return Integer.compare(a.y, b.y);
                }
                return Integer.compare(a.x, b.x);
            });

            for (int i = 1; i < n; i++) {
                if (tab[i].y < tab[i - 1].y) {
                    out.println(ans);
                    return;
                }
            }
            tosub = 1;
            for (int i = 0; i < n; i++) {
                int act = i;
                while (act < n && tab[act].x == tab[i].x && tab[act].y == tab[i].y) act++;
                act--;
                if (act - i > 0) {
                    tosub *= fact[act - i + 1];
                    tosub %= mod;
                }
                i = act;
            }
            ans += tosub;
            ans %= mod;
//        Arrays.sort(tab, (a, b) -> {
//            if (a.y == b.y) {
//                return compare(a.x, b.x);
//            }
//            return compare(a.y,b.y);
//        });
//        tosub=1;
//        for (int i=0;i<n;i++){
//            int act=i;
//            while (act<n && tab[act].y==tab[i].y) act++;
//            act--;
//            for (int j=i;j<=act;j++){
//                int y=j;
//                int s=j;
//                while (y<=act && tab[y].x==tab[s].x) y++;
//                y--;
//                if (y-s>0){
//                    if (tosub==0) {
//                        tosub=fact[y-s+1];
//                    }
//                    else{
//                        tosub*=fact[y-s+1];
//                        tosub%=mod;
//                    }
//                }
//                j=y;
//            }
//            i=act;
//        }
//        ans+=tosub;
            out.println(ans);
        }

        static class Pair {
            int x;
            int y;

            public Pair(int x, int y) {
                this.x = x;
                this.y = y;
            }

        }

    }

    static class inputClass {
        BufferedReader br;
        StringTokenizer st;

        public inputClass(InputStream in) {

            br = new BufferedReader(new InputStreamReader(in));
        }

        public String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

    }
}

