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. 
#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
struct P {
  int a, b;
  bool operator<(const P &o) const { return a == o.a ? b < o.b : a < o.a; }
};
P s[300005];
long long fac[300005], as[300005], bs[300005];
int main() {
  int N;
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    scanf("%d %d", &s[i].a, &s[i].b);
    as[i] = s[i].a;
    bs[i] = s[i].b;
  }
  fac[1] = 1;
  for (int i = 2; i <= N; i++) fac[i] = (fac[i - 1] * i) % mod;
  sort(as + 1, as + N + 1);
  sort(bs + 1, bs + N + 1);
  sort(s + 1, s + N + 1);
  long long ans = fac[N];
  long long l = 1, asort = 1, bsort = 1, csort = 1;
  for (int i = 2; i <= N; i++)
    if (as[i] != as[l]) {
      asort = (asort * fac[i - l]) % mod;
      l = i;
    }
  asort = (asort * fac[N - l + 1]) % mod;
  l = 1;
  for (int i = 2; i <= N; i++)
    if (bs[i] != bs[l]) {
      bsort = (bsort * fac[i - l]) % mod;
      l = i;
    }
  bsort = (bsort * fac[N - l + 1]) % mod;
  l = 1;
  for (int i = 2; i <= N; i++) {
    if (s[i].a < s[i - 1].a || s[i].b < s[i - 1].b) {
      csort = 0;
      break;
    }
    if (s[i].a != s[l].a || s[i].b != s[l].b) {
      csort = (csort * fac[i - l]) % mod;
      l = i;
    }
  }
  csort = (csort * fac[N - l + 1]) % mod;
  printf("%lld\n", (ans - asort - bsort + csort + 2 * mod) % mod);
  return 0;
}
