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 pt {
  int a, b;
};
bool cmp1(pt a, pt b) { return a.a < b.a; }
bool cmp2(pt a, pt b) { return a.b < b.b; }
bool cmp12(pt a, pt b) { return a.a < b.a && a.b < b.b; }
long long fact(int n) {
  long long res = 1;
  for (int i = 2; i <= n; ++i) res = res * i % mod;
  return res;
}
int main() {
  int n, cnt;
  cin >> n;
  vector<pt> s(n);
  for (int i = 0; i < n; ++i) cin >> s[i].a >> s[i].b;
  sort(s.begin(), s.end(), &cmp1);
  long long ans1 = 1;
  cnt = 1;
  for (int i = 1; i < n; ++i) {
    if (s[i].a != s[i - 1].a) {
      ans1 = ans1 * fact(cnt) % mod;
      cnt = 1;
    } else
      ++cnt;
  }
  ans1 = ans1 * fact(cnt) % mod;
  sort(s.begin(), s.end(), &cmp2);
  long long ans2 = 1;
  cnt = 1;
  for (int i = 1; i < n; ++i) {
    if (s[i].b != s[i - 1].b) {
      ans2 = ans2 * fact(cnt) % mod;
      cnt = 1;
    } else
      ++cnt;
  }
  ans2 = ans2 * fact(cnt) % mod;
  sort(s.begin(), s.end(), &cmp12);
  long long ans12 = 1;
  for (int i = 1; i < n; ++i)
    ans12 *= (s[i].a >= s[i - 1].a && s[i].b >= s[i - 1].b);
  if (ans12) {
    cnt = 1;
    for (int i = 1; i < n; ++i) {
      if (s[i].a != s[i - 1].a || s[i].b != s[i - 1].b) {
        ans12 = ans12 * fact(cnt) % mod;
        cnt = 1;
      } else
        ++cnt;
    }
    ans12 = ans12 * fact(cnt) % mod;
  }
  long long ans = fact(n);
  cout << (((ans - ans1 + mod) % mod - ans2 + mod) % mod + ans12) % mod;
  return 0;
}
