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. 
from operator import itemgetter
import sys


class Combination:
    '''MOD上の
    計算量：階乗・逆元テーブルの作成O(N)
    nCkを求めるO(1)'''

    def __init__(self, n, MOD):
        self.fact = [1]
        for i in range(1, n + 1):
            self.fact.append(self.fact[-1] * i % MOD)


input = sys.stdin.readline
n = int(input())
p = [list(map(int, input().split())) for i in range(n)]
MOD = 998244353
comb = Combination(n, MOD)
memo1 = {}
memo2 = {}

for i in range(n):
    tmp1, tmp2 = p[i]
    if tmp1 not in memo1:
        memo1[tmp1] = 1
    else:
        memo1[tmp1] += 1
    if tmp2 not in memo2:
        memo2[tmp2] = 1
    else:
        memo2[tmp2] += 1

ans = 0

tmp = 1
for i in memo1:
    tmp *= comb.fact[memo1[i]]
    tmp %= MOD
ans += tmp
ans %= MOD

tmp = 1
for i in memo2:
    tmp *= comb.fact[memo2[i]]
    tmp %= MOD
ans += tmp
ans %= MOD

p = sorted(p, key = itemgetter(1))
p = sorted(p, key = itemgetter(0))
for i in range(n-1):
    if p[i][1] > p[i+1][1]:
        ans = comb.fact[n] - ans
        ans %= MOD
        print(ans)
        exit()
cnt = 1
tmp = 1
for i in range(n-1):
    if p[i][0] == p[i+1][0] and p[i][1] == p[i+1][1]:
        cnt += 1
    else:
        tmp *= comb.fact[cnt]
        tmp %= MOD
        cnt = 1
tmp *= comb.fact[cnt]
tmp %= MOD
ans -= tmp

ans = comb.fact[n] - ans
ans %= MOD
print(ans)