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 OMKARHELPME{
private static final long mod = 998244353;
private static Map<Long, Long> aVals = new HashMap<Long, Long>();
private static Map<Pair, Long> pairVals = new HashMap<Pair, Long>();
public static void main(String[] args)
{
   Scanner scanscan = new Scanner(System.in);
   int size = scanscan.nextInt();
   long[] a = new long[size];
   long[] b = new long[size];
   Pair[] pairs = new Pair[size];
   long x, y;
   for(int i = 0; i < size; i++)
   {
      x = scanscan.nextLong();
      y = scanscan.nextLong();
      a[i] = x;
      b[i] = y;
      pairs[i] = new Pair(x, y);
   }
   long ans = factorial(size) - bad(a) - bad(b) + bothBad(pairs);
   ans = ans % mod;
   while(ans >= mod)
   {
      ans -= mod;
   }
   while(ans < 0)
   {
      ans += mod;
   }
   System.out.println(""+ans);
   }
public static long bad(long[] arr)
{
   Map<Long, Long> vals = new HashMap<Long, Long>();
   for(int i = 0; i < arr.length; i++)
   {
      if(!vals.containsKey(arr[i]))
      {
         vals.put(arr[i], (long)(1));
      }
      else
      {
         vals.put(arr[i], vals.get(arr[i]) + 1);
      }
   }
   Object[] temp = vals.keySet().toArray();
   long permutations = 1;
   for(int i = 0; i < temp.length; i++)
   {
      permutations = (factorial(vals.get(temp[i]))*permutations) % mod;
   }
   return permutations;
}
public static long bothBad(Pair[] arr)
{
   Arrays.sort(arr);
   for(int i = 0; i < arr.length-1; i++)
   {
      if(arr[i].b > arr[i+1].b)
      {
         return 0;
      }
   }
   long permutations = 1;
   long present = 1;
   for(int i = 0; i < arr.length-1; i++)
   {
      if(arr[i].a == arr[i+1].a && arr[i].b == arr[i+1].b)
      {
         present += 1;
      }
      else
      {
         present = 1;
      }
      permutations = (permutations * present) % mod;
   }
   return permutations;
}
public static long factorial(long a)
{
   long ans = 1;
   for(int i = 1; i <= a; i++)
   {
      ans = (ans * i) % mod;
   }
   return ans;
}
static class Pair implements Comparable<Pair> {
final int a;
final int b;
Pair(long a, long b) {
this.a = (int)(a);
this.b = (int)(b);
}
@Override
public int compareTo(Pair other) {
if (a != other.a) {
return (a - other.a);
} else {
return (b - other.b);
}}}
}

