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;
bool comp(pair<long long int, long long int> p1,
          pair<long long int, long long int> p2) {
  if ((p1.first <= p2.first) && (p1.second <= p2.second))
    return true;
  else
    return false;
}
int main() {
  long long int n, i, a, b, s = 1, z1 = 1, mod = 998244353, neg = 0, ans = 0,
                            ans1 = 0, f = 0, z2 = 1;
  cin >> n;
  long long int fact[n + 1];
  vector<long long int> vec, vec1;
  vector<pair<long long int, long long int> > vec2;
  for (i = 1; i <= n; i++) {
    cin >> a >> b;
    vec.push_back(a);
    vec1.push_back(b);
    vec2.push_back({a, b});
    if (i == 1)
      fact[1] = 1;
    else
      fact[i] = ((fact[i - 1] % mod) * (i % mod)) % mod;
  }
  sort(vec.begin(), vec.end());
  sort(vec1.begin(), vec1.end());
  sort(vec2.begin(), vec2.end());
  auto it = vec.begin();
  while (it < vec.end()) {
    auto it1 = vec.begin();
    for (it1 = it; it1 != vec.end(); it1++) {
      if ((*it) != (*it1)) {
        break;
      }
    }
    long long int z;
    z = (it1 - it);
    s = ((s % mod) * (fact[z] % mod)) % mod;
    if (it1 == vec.end()) {
      break;
    }
    it = it1;
  }
  it = vec1.begin();
  while (it < vec1.end()) {
    auto it1 = vec1.begin();
    for (it1 = it; it1 != vec1.end(); it1++) {
      if ((*it) != (*it1)) {
        break;
      }
    }
    long long int z;
    z = (it1 - it);
    z1 = ((z1 % mod) * (fact[z] % mod)) % mod;
    if (it1 == vec1.end()) {
      break;
    }
    it = it1;
  }
  ans1 = ((ans1 % mod + s) % mod + z1 % mod) % mod;
  auto it2 = vec2.begin();
  for (it2 = vec2.begin() + 1; it2 != vec2.end(); it2++) {
    auto it1 = it2 - 1;
    if (((it2->first) < (it1->first)) || ((it2->second) < (it1->second))) {
      f = 1;
      break;
    }
  }
  if (f == 0) {
    it2 = vec2.begin();
    while (it2 < vec2.end()) {
      auto it1 = vec2.begin();
      for (it1 = it2; it1 != vec2.end(); it1++) {
        if (((it2->first) != (it1->first)) ||
            ((it2->second) != (it1->second))) {
          break;
        }
      }
      long long int z;
      z = (it1 - it2);
      z2 = ((z2 % mod) * (fact[z] % mod)) % mod;
      if (it1 == vec2.end()) {
        break;
      }
      it2 = it1;
    }
  } else
    z2 = 0;
  ans1 = (ans1 - z2 + mod) % mod;
  ans = (fact[n] - ans1 + mod) % mod;
  cout << ans;
}
