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;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
bool sortinrev(const pair<int, int>& a, const pair<int, int>& b) {
  if (a.first == b.first) return b.second > a.second;
  return a.first < b.first;
}
void run(vector<int> v) {
  for (int i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << endl;
}
int modInverse(long long int a, long long int m) {
  long long int m0 = m;
  long long int y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long int q = a / m;
    long long int t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
void numberOfNodes(int n, int s, int e, vector<int>& count1,
                   vector<int> adj[]) {
  vector<int>::iterator u;
  count1[s] = 1;
  for (u = adj[s].begin(); u != adj[s].end(); u++) {
    if (*u == e) continue;
    numberOfNodes(n, *u, s, count1, adj);
    count1[s] += count1[*u];
  }
}
void solve() {
  int n;
  cin >> n;
  vector<long long int> fact(n + 1);
  fact[0] = 1;
  for (int i = 1; i <= n; i++) {
    fact[i] = (fact[i - 1] * i) % 998244353;
  }
  vector<pair<long long int, long long int>> v(n);
  unordered_map<int, int> mp1, mp2;
  long long int ans = 0, ans1 = 1, ans2 = 1, ans3 = 1;
  vector<int> a(n), b(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i].first >> v[i].second;
    a[i] = v[i].first;
    b[i] = v[i].second;
    mp1[a[i]]++;
    mp2[b[i]]++;
  }
  sort(a.begin(), a.end());
  sort(b.begin(), b.end());
  sort(v.begin(), v.end(), sortinrev);
  int breaking = 0;
  for (int i = 0; i < n - 1; i++) {
    if (v[i].second > v[i + 1].second) {
      breaking = 1;
      break;
    }
  }
  for (auto x : mp1) {
    ans1 = (ans1 * fact[x.second]) % 998244353;
    ans1 = ans1 % 998244353;
  }
  for (auto x : mp2) {
    ans2 = (ans2 * fact[x.second]) % 998244353;
    ans2 = ans2 % 998244353;
  }
  if (breaking == 0) {
    int x;
    for (int i = 0; i < n - 1; i++) {
      x = 1;
      if ((v[i].second == v[i + 1].second) && (v[i].first == v[i + 1].first)) {
        while ((i + 1 < n) && (v[i].second == v[i + 1].second) &&
               (v[i].first == v[i + 1].first)) {
          ++x;
          ++i;
        }
      }
      ans3 = (ans3 * fact[x]) % 998244353;
      ans3 = ans3 % 998244353;
    }
  } else
    ans3 = 0;
  cout << (((fact[n] - (ans1 + ans2 - ans3)) % 998244353) + 998244353) %
              998244353;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
