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;
using ll = long long;
template <class T, class U>
ostream& operator<<(ostream& o, const pair<T, U>& p) {
  o << "(" << p.first << "," << p.second << ")";
  return o;
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& v) {
  o << "[";
  for (T t : v) {
    o << t << ",";
  }
  o << "]";
  return o;
}
const int N = 300003;
const ll mod = 998244353;
ll f[N];
int main() {
  f[0] = 1;
  for (int i = 1; i < N; ++i) f[i] = (f[i - 1] * i) % mod;
  int n;
  scanf(" %d", &n);
  vector<int> a(n), b(n);
  for (int i = 0; i < (int)(n); ++i) {
    scanf(" %d %d", &a[i], &b[i]);
    --a[i];
    --b[i];
  }
  auto one_bad = [&](const vector<int>& v) {
    vector<int> ct(n);
    for (int i : v) ++ct[i];
    ll ret = 1;
    for (int i = 0; i < (int)(n); ++i) (ret *= f[ct[i]]) %= mod;
    return ret;
  };
  ll ans = f[n];
  ans = (ans - one_bad(a) + mod) % mod;
  ans = (ans - one_bad(b) + mod) % mod;
  vector<vector<int>> B(n);
  for (int i = 0; i < (int)(n); ++i) B[a[i]].push_back(b[i]);
  ll both = 1;
  int prev_max = -1;
  for (int i = 0; i < (int)(n); ++i) {
    if (B[i].empty()) continue;
    int mn = n, mx = -1;
    map<int, int> ct;
    for (int j : B[i]) {
      ++ct[j];
      mn = min(mn, j);
      mx = max(mx, j);
    }
    if (prev_max > mn) {
      both = 0;
      break;
    }
    for (const auto& p : ct) (both *= f[p.second]) %= mod;
    prev_max = mx;
  }
  (ans += both) %= mod;
  printf("%lld\n", ans);
  return 0;
}
