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 M = 998244353;
long long f[300005];
vector<long long> m1[300005], m2[300005];
int32_t main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  ;
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    m1[x].push_back(y);
    m2[y].push_back(x);
  }
  f[0] = 1;
  for (long long i = 1; i < 300005; i++) f[i] = f[i - 1] * i % M;
  long long ans = f[n], d1 = 1, d2 = 1, d3 = 1;
  for (long long i = 1; i <= n; i++) d1 = d1 * f[m1[i].size()] % M;
  for (long long i = 1; i <= n; i++) d2 = d2 * f[m2[i].size()] % M;
  ans = ((ans - d1 - d2) % M + M) % M;
  vector<long long> a, b;
  for (long long i = 1; i <= n; i++) {
    if (m1[i].size() == 0) continue;
    sort((m1[i]).begin(), (m1[i]).end());
    map<long long, long long> m;
    for (long long e : m1[i]) m[e]++;
    for (pair<long long, long long> p : m) d3 = d3 * f[p.second] % M;
    a.push_back(i);
    b.push_back(m1[i][0]);
    b.push_back(m1[i].back());
  }
  for (long long i = 0; i < a.size() - 1; i++)
    if (a[i + 1] < a[i]) d3 = 0;
  for (long long i = 0; i < b.size() - 1; i++)
    if (b[i + 1] < b[i]) d3 = 0;
  ans = (ans + d3) % M;
  cout << ans << '\n';
}
