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 int mod = 998244353, N = 3e5 + 5;
long long fact[N];
template <class T>
long long calc(vector<T> v) {
  int cnt = 0;
  T last = v[0];
  long long ans = 1;
  for (int i = 0; i < (int)v.size(); ++i) {
    if (last != v[i]) {
      ans = ans * fact[cnt] % mod;
      cnt = 0;
      last = v[i];
    }
    ++cnt;
  }
  ans = ans * fact[cnt] % mod;
  if (ans < 0) ans += mod;
  return ans;
}
int main() {
  fact[0] = 1;
  for (int i = 1; i < N; ++i) {
    fact[i] = 1LL * i * fact[i - 1] % mod;
  }
  int n;
  scanf("%d", &n);
  vector<int> a(n), b(n);
  vector<pair<int, int>> s(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d %d", &a[i], &b[i]);
    s[i] = {a[i], b[i]};
  }
  sort(a.begin(), a.end());
  sort(b.begin(), b.end());
  sort(s.begin(), s.end());
  long long ans = fact[n] - calc(a) - calc(b);
  bool ok = 1;
  for (int i = 0; i + 1 < n; ++i) {
    if (s[i].second > s[i + 1].second) {
      ok = 0;
    }
  }
  if (ok) ans += calc(s);
  ans %= mod;
  if (ans < 0) ans += mod;
  printf("%lld\n", ans);
  return 0;
}
