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;
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int>> ps(n);
  for (int i = 0; i < n; i++) {
    cin >> ps[i].first >> ps[i].second;
  }
  sort(ps.begin(), ps.end());
  vector<int> a(n), b(n);
  for (int i = 0; i < n; i++) {
    a[i] = ps[i].first;
    b[i] = ps[i].second;
  }
  vector<long long> fs(n + 1);
  fs[0] = 1;
  for (int i = 1; i <= n; i++) {
    fs[i] = fs[i - 1] * i % MOD;
  }
  long long bad = 1;
  for (int i = 0, j; i < n; i = j) {
    for (j = i + 1; j < n; j++) {
      if (a[j] != a[i]) {
        break;
      }
    }
    bad = bad * fs[j - i] % MOD;
  }
  bool is_zigzag = !is_sorted(b.begin(), b.end());
  long long bad1 = 1;
  sort(b.begin(), b.end());
  for (int i = 0, j; i < n; i = j) {
    for (j = i + 1; j < n; j++) {
      if (b[j] != b[i]) {
        break;
      }
    }
    bad1 = bad1 * fs[j - i] % MOD;
  }
  if (is_zigzag) {
    bad = (bad + bad1) % MOD;
  } else {
    long long bad2 = 1;
    for (int i = 0, j; i < n; i = j) {
      for (j = i + 1; j < n; j++) {
        if (b[j] != b[i]) {
          break;
        }
      }
      for (int ii = i, jj; ii < j; ii = jj) {
        for (jj = ii + 1; jj < j; jj++) {
          if (a[jj] != a[ii]) {
            break;
          }
        }
        bad2 = bad2 * fs[jj - ii] % MOD;
      }
    }
    bad = (bad + bad1 - bad2 + MOD) % MOD;
  }
  long long res = (fs[n] - bad + MOD) % MOD;
  cout << res << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  solve();
}
