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

in_file = sys.stdin#

n = int(in_file.readline().strip())
Al = []
Ar = []
A = []
for _ in range(n):
    x, y = map(int, in_file.readline().strip().split())
    Al.append(x)
    Ar.append(y)
    A.append((x,y))

Al.sort()
Ar.sort()
A.sort()

def groups(L):
    p = L[0]
    G = [1]
    for i in range(1, len(L)):
        if p == L[i]:
            G[-1] += 1
        else:
            p = L[i]
            G.append(1)
    return G

def bin_exp(x, a, p):
    r = 1
    while a>0:
        if a%2 == 1:
            r = (r*x)%p
        x = (x*x)%p
        a = a//2
    return r

def mod_fact(x, p):
    r = 1
    for i in range(1, x+1):
        r = (r*i)%p
    return r

p = 998244353
Gl = groups(Al)
rl = 1
for i in Gl:
    rl = (rl*mod_fact(i, p))%p
Gr = groups(Ar)
rr = 1
for i in Gr:
    rr = (rr*mod_fact(i, p))%p
G = groups(A)
r = 1
for i in G:
    r = (r*mod_fact(i, p))%p
if list(zip(Al, Ar)) != A:
    r = 0
res = (mod_fact(len(A), p)+r)%p
res = (p+res-rl)%p
res = (p+res-rr)%p

sys.stdout.write(str(res))
sys.stdout.flush() 