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 sys

mod = 998244353

input = sys.stdin.readline
n = int(input())
s = [list(map(int, input().split())) for y in range(n)]

 
fact = [1]*(n+1)                    # storing factorials for further use
for i in range(1, n+1):
    fact[i] = (fact[i-1] * i) % mod
  

a = [0]*n                           # first elements
b = [0]*n                           # second elements

for x, y in s:                      
    a[x-1] += 1                     # counting value repetitions for first elements
    b[y-1] += 1                     # counting value repetitions for second elements
 
a_count = 1                         # number of first elements sorted permutations
b_count = 1                         # number of second elements sorted permutations
for i in range(n):                  # product of sub-permutations of repeated values in first elements
    a_count *= fact[a[i]]
    a_count %= mod
    b_count *= fact[b[i]]           # product of sub-permutations of repeated values in first elements
    b_count %= mod
 
ab_count = 1                        # number of both elements sorted permutations
count = 1                           # current tuple repetition´s count
s.sort()                            # tuples default sorting
for i in range(1, n):
    if s[i][1] < s[i-1][1]:         # if second elements are not sorted there are no permutations in which both tuple elements are sorted
        ab_count = 0
        break
 
    if s[i] == s[i-1]:              # being s sorted, one knows that a repeated tuple would be found comparing with the predecessor of the current one
        count += 1
    else:                           # found a different tuple, update counters     
        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)
