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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.StringTokenizer;

public class NumberOfPermutations {
    public static final int MOD = 998244353;
    public static void main(String[] args)throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(in.readLine());
        int N = Integer.parseInt(st.nextToken());
        int[][] pairs = new int[N][2];
        HashMap<Integer, Integer> countA = new HashMap<>();
        HashMap<Integer, Integer> countB = new HashMap<>();
        HashMap<Integer, HashMap<Integer, Integer>> countAAndB = new HashMap<>();
        for (int i = 0; i < N; i++) {
            st = new StringTokenizer(in.readLine());
            int a = Integer.parseInt(st.nextToken());
            countA.putIfAbsent(a, 0);
            countA.replace(a, countA.get(a)+1);
            int b = Integer.parseInt(st.nextToken());
            countB.putIfAbsent(b, 0);
            countB.replace(b, countB.get(b) + 1);
            countAAndB.putIfAbsent(a, new HashMap<>());
            countAAndB.get(a).putIfAbsent(b, 0);
            countAAndB.get(a).replace(b, countAAndB.get(a).get(b)+1);
            pairs[i][0] = a;
            pairs[i][1] = b;
        }
        Arrays.sort(pairs, new Comparator<int[]>() {
            @Override
            public int compare(int[] ints, int[] t1) {
                if (ints[0] != t1[0]) {
                    return Integer.compare(ints[0], t1[0]);
                }
                else {
                    return (Integer.compare(ints[1], t1[1]));
                }
            }
        });
        long goodA = 1;
        long goodB = 1;
        for (int key : countA.keySet()) {
            goodA *= factorial(countA.get(key));
            goodA %= MOD;
            assert goodA > 0;
        }
        for (int key : countB.keySet()) {
            goodB *= factorial(countB.get(key));
            goodB %= MOD;
            assert goodB > 0;
        }
//        int last = 0;
//        int curA = 0;
//        HashMap<Integer, Integer> count = new HashMap<>();
//        int total = 1;
//        for (int i = 0; i < N; i++) {
//            if (curA != pairs[i][0]) {
//                int totalRearrange = 1;
//                curA = pairs[i][0];
//                for (int keys : count.keySet()) {
//                    totalRearrange *= factorial(count.get(keys));
//                    assert totalRearrange > 0;
//                }
//                total *= totalRearrange;
//                assert total > 0;
////                total %= MOD;
//                count = new HashMap<>();
//            }
//            if (last <= pairs[i][1]) {
//                last = pairs[i][1];
//                count.putIfAbsent(pairs[i][1], 0);
//                count.replace(pairs[i][1], count.get(pairs[i][1]) + 1);
//            }
//            else {
//                total = 0;
//                break;
//            }
//        }
//        if (total != 0) {
//            int totalRearrange = 1;
//            for (int keys : count.keySet()) {
//                totalRearrange *= factorial(count.get(keys));
//                totalRearrange %= MOD;
//            }
//            total *= totalRearrange;
//            total %= MOD;
//        }
        long total = 1;
        for (int key : countAAndB.keySet()) {
            for (int nextKey : countAAndB.get(key).keySet()) {
                total *= factorial(countAAndB.get(key).get(nextKey));
                total %= MOD;
            }
        }
        for (int j = 1; j < N; j++) {
            if (pairs[j-1][1] > pairs[j][1]) {
                total = 0;
                break;
            }
        }
        long ans = factorial(N) + total - goodA - goodB + 2 * MOD;
        assert ans > 0;
        ans %= MOD;
        System.out.println(ans);
    }
    public static long factorial(int n) {
        long sum = 1;
        while (n > 0) {
            sum *= n;
            sum %= MOD;
            n--;
        }
        return sum;
    }
}
