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.*;
import java.util.function.Function;

public class Main {

    public static void main(String[] args) {

        Parser parser = new Parser();
        int t = 1;
        for (int i = 1; i <= t; i++) {
            Problem problem = new Problem();
            problem.solve(i);
        }
    }
}

class Problem {
    Parser parser = new Parser();
    private static final long MOD = 998244353;

    public void solve(int testNum) {

        int n = parser.parseInt();
        List<Pair> arr = new ArrayList<>();

        long[] fact = new long[n + 1];
        fact[0] = 1;
        for (int i = 1; i <= n; i++) {
            fact[i] = (fact[i - 1] * i) % MOD;
        }

        int[] aCount = new int[n + 1];
        int[] bCount = new int[n + 1];
        for (int i = 0; i < n; i++) {
            int l = parser.parseInt();
            int r = parser.parseInt();
            aCount[l] += 1;
            bCount[r] += 1;
            arr.add(new Pair(l, r));
        }

        arr.sort(Comparator.comparingInt(Pair::getL)
                .thenComparingInt(Pair::getR));

        long aV = 1;
        for (int i = 1; i <= n; i++) {
            aV = (aV * fact[aCount[i]]) % MOD;
        }

        long bV = 1;
        for (int i = 1; i <= n; i++) {
            bV = (bV * fact[bCount[i]]) % MOD;
        }

        for(int i = 1; i < n; i++) {
            Pair a = arr.get(i - 1);
            Pair b = arr.get(i);
            if(a.r > b.r) {
                System.out.println((fact[n] - (aV + bV) + 3 * MOD) % MOD);
                return;
            }
        }

        Map<Pair, Integer> counts = new HashMap<>();
        for(int i = 0; i < n; i++) {
            Pair pair = arr.get(i);
            counts.put(pair, counts.getOrDefault(pair, 0) + 1);
        }

        long cV = 1;
        for(Map.Entry<Pair, Integer> entry : counts.entrySet()) {
            cV = (cV * fact[entry.getValue()]) % MOD;
        }

        System.out.println((fact[n] - (aV + bV - cV) + 3 * MOD) % MOD);
    }

    static class Pair {
        int l, r;

        public Pair(int l, int r) {
            this.l = l;
            this.r = r;
        }

        public int getL() {
            return l;
        }

        public int getR() {
            return r;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Pair pair = (Pair) o;
            return l == pair.l &&
                    r == pair.r;
        }

        @Override
        public int hashCode() {
            return Objects.hash(l, r);
        }

        @Override
        public String toString() {
            return "Pair{" +
                    "l=" + l +
                    ", r=" + r +
                    '}';
        }
    }
}

class MathUtil {

    public static List<Integer> getPrimes(int r) {
        if (r < 2) {
            return List.of();
        }

        List<Integer> result = new ArrayList<>();
        result.add(2);
        boolean[] is = new boolean[r + 1];
        for (int i = 4; i <= r; i += 2) {
            is[i] = true;
        }
        for (int i = 3; i <= r; i += 2) {
            if (is[i]) continue;
            result.add(i);
            for (long j = (long) i * i; j <= r; j += i) {
                is[(int) j] = true;
            }
        }
        return result;
    }

    public static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    public static long pow(long x, long y) {
        return pow(x, y, Function.identity());
    }

    public static long pow(long x, long y, long mod) {
        return pow(x, y, v -> v % mod);
    }

    private static long pow(long x, long y, Function<Long, Long> action) {
        long res = 1;
        long t = x;
        while (y > 0) {
            if ((y & 1) == 1) {
                res += t;
                res = action.apply(res);
            }
            t *= t;
            t = action.apply(t);
            y /= 2;
        }
        return res;
    }
}

class DisjointSet {

    private final int[] f;

    public DisjointSet(int size) {
        f = new int[size + 1];
        for (int i = 0; i <= size; i++) {
            f[i] = i;
        }
    }

    public int find(int u) {
        if (f[u] == u) return u;
        return f[u] = find(f[u]);
    }

    public boolean merge(int u, int v) {
        u = find(u);
        v = find(v);
        if (u == v) return false;
        if (u > v) {
            int t = u;
            u = v;
            v = t;
        }
        f[u] = v;
        return true;
    }
}

class Parser {
    private static final BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    private StringTokenizer tokenizer;

    public int parseInt() {
        fill();
        return Integer.parseInt(tokenizer.nextToken());
    }

    public long parseLong() {
        fill();
        return Long.parseLong(tokenizer.nextToken());
    }

    public double parseDouble() {
        fill();
        return Double.parseDouble(tokenizer.nextToken());
    }

    public String parseLine() {
        try {
            return reader.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return null;
    }

    String parseString() {
        fill();
        return tokenizer.nextToken();
    }

    private void fill() {
        while (tokenizer == null || !tokenizer.hasMoreElements()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
    }
}