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;
int64_t inf = 1000 * 1000 * 1000;
int64_t inf64 = inf * inf;
int64_t mod = 998244353;
int64_t nll = 0;
int64_t cnt[300005][3];
map<pair<int32_t, int32_t>, int32_t> mp;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cerr << fixed << setprecision(8);
  cout << fixed << setprecision(8);
  srand(time(0));
  auto START_TIME = chrono::high_resolution_clock::now();
  double START_TIME_clock = clock();
  int32_t n;
  cin >> n;
  vector<pair<int64_t, int64_t> > v(n);
  for (int32_t i = 0; i < n; ++i) {
    cin >> v[i].first >> v[i].second;
    ++mp[v[i]];
    ++cnt[v[i].first][1];
    ++cnt[v[i].second][2];
  }
  int64_t ans1 = 1, ans2 = 1, ans3 = 1;
  for (int32_t i = 0; i < n; ++i) {
    if (mp[v[i]] > 0) {
      int64_t temp = mp[v[i]];
      while (temp) {
        ans3 = (ans3 * temp) % mod;
        --temp;
      }
      mp[v[i]] = 0;
    }
  }
  sort(v.begin(), v.end());
  bool good = true;
  for (int32_t i = 0; i < n - 1; ++i) {
    if (v[i].first > v[i + 1].first || v[i].second > v[i + 1].second)
      good = false;
  }
  if (!good) ans3 = 0;
  for (int32_t i = 0; i <= n; ++i) {
    if (cnt[i][1] > 0) {
      int64_t temp = cnt[i][1];
      while (temp) {
        ans1 = (ans1 * temp) % mod;
        --temp;
      }
    }
    if (cnt[i][2] > 0) {
      int64_t temp = cnt[i][2];
      while (temp) {
        ans2 = (ans2 * temp) % mod;
        --temp;
      }
    }
  }
  int64_t ans = 1;
  for (int32_t i = 1; i <= n; ++i) ans = (ans * i) % mod;
  ans -= ans1;
  ans -= ans2;
  ans += ans3;
  ans += mod * mod;
  ans %= mod;
  cout << ans;
  cerr << endl << (clock() - START_TIME_clock) / CLOCKS_PER_SEC << " sec.";
  cerr << endl
       << chrono::duration<long double>(chrono::high_resolution_clock::now() -
                                        START_TIME)
              .count()
       << " sec.";
}
