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. 
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------

import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
import threading



#sys.setrecursionlimit(300000)
#threading.stack_size(10**8)


BUFSIZE = 8192
 
 
class FastIO(IOBase):
    newlines = 0
    
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
     
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
  
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

#-------------------game starts now-----------------------------------------------------
class Factorial:
    def __init__(self, MOD):
        self.MOD = MOD
        self.factorials = [1, 1]
        self.invModulos = [0, 1]
        self.invFactorial_ = [1, 1]
 
    def calc(self, n):
        if n <= -1:
            print("Invalid argument to calculate n!")
            print("n must be non-negative value. But the argument was " + str(n))
            exit()
        if n < len(self.factorials):
            return self.factorials[n]
        nextArr = [0] * (n + 1 - len(self.factorials))
        initialI = len(self.factorials)
        prev = self.factorials[-1]
        m = self.MOD
        for i in range(initialI, n + 1):
            prev = nextArr[i - initialI] = prev * i % m
        self.factorials += nextArr
        return self.factorials[n]
 
    def inv(self, n):
        if n <= -1:
            print("Invalid argument to calculate n^(-1)")
            print("n must be non-negative value. But the argument was " + str(n))
            exit()
        p = self.MOD
        pi = n % p
        if pi < len(self.invModulos):
            return self.invModulos[pi]
        nextArr = [0] * (n + 1 - len(self.invModulos))
        initialI = len(self.invModulos)
        for i in range(initialI, min(p, n + 1)):
            next = -self.invModulos[p % i] * (p // i) % p
            self.invModulos.append(next)
        return self.invModulos[pi]
 
    def invFactorial(self, n):
        if n <= -1:
            print("Invalid argument to calculate (n^(-1))!")
            print("n must be non-negative value. But the argument was " + str(n))
            exit()
        if n < len(self.invFactorial_):
            return self.invFactorial_[n]
        self.inv(n)  # To make sure already calculated n^-1
        nextArr = [0] * (n + 1 - len(self.invFactorial_))
        initialI = len(self.invFactorial_)
        prev = self.invFactorial_[-1]
        p = self.MOD
        for i in range(initialI, n + 1):
            prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
        self.invFactorial_ += nextArr
        return self.invFactorial_[n]
 
 
class Combination:
    def __init__(self, MOD):
        self.MOD = MOD
        self.factorial = Factorial(MOD)
 
    def ncr(self, n, k):
        if k < 0 or n < k:
            return 0
        k = min(k, n - k)
        f = self.factorial
        return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
#-------------------------------------------------------------------------
n = int(input())
s = [list(map(int, input().split())) for y in range(n)]
MOD = 998244353
 
fact = [1]*(n+1)
for i in range(1, n+1):
    fact[i] = (fact[i-1] * i) % MOD
 
a, b = [0]*n, [0]*n
 
for x, y in s:
    a[x-1] += 1
    b[y-1] += 1
 
a_count = 1
b_count = 1
for i in range(n):
    a_count *= fact[a[i]]
    a_count %= MOD
    b_count *= fact[b[i]]
    b_count %= MOD
 
ab_count = 1
count = 1
s.sort()
for i in range(1, n):
    if s[i][1] < s[i-1][1]:
        ab_count = 0
        break
 
    if s[i] == s[i-1]:
        count += 1
    else:
        ab_count *= fact[count]
        ab_count %= MOD
        count = 1
 
ab_count *= fact[count]
ab_count %= MOD
 
print((fact[n] - a_count - b_count + ab_count) % MOD)