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 MAX = 300006;
const long long MOD = 998244353LL;
long long fact[MAX] = {};
vector<pair<int, int> > v;
vector<int> v1, v2;
bool operator<(pair<int, int> x, pair<int, int> y) {
  if (x.first == y.first) {
    return x.second < y.second;
  }
  return x.first < y.first;
}
bool operator==(pair<int, int> x, pair<int, int> y) {
  return (x.first == y.first && x.second == y.second);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  fact[0] = 1LL;
  for (int i = 1; i < MAX; ++i) {
    fact[i] = fact[i - 1] * i;
    fact[i] %= MOD;
  }
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    pair<int, int> x{a, b};
    v.push_back(x);
    v1.push_back(a);
    v2.push_back(b);
  }
  long long tot = fact[n], cnt1 = 1LL, cnt2 = 1LL, cnt12 = 0LL;
  sort(v.begin(), v.end());
  sort(v1.begin(), v1.end());
  sort(v2.begin(), v2.end());
  long long di = 1LL;
  for (int i = 1; i < v1.size(); ++i) {
    if (v1[i] != v1[i - 1]) {
      cnt1 *= fact[di];
      cnt1 %= MOD;
      di = 1LL;
    } else {
      ++di;
    }
  }
  cnt1 *= fact[di];
  cnt1 %= MOD;
  di = 1LL;
  for (int i = 1; i < v2.size(); ++i) {
    if (v2[i] != v2[i - 1]) {
      cnt2 *= fact[di];
      cnt2 %= MOD;
      di = 1LL;
    } else {
      ++di;
    }
  }
  cnt2 *= fact[di];
  cnt2 %= MOD;
  di = 1LL;
  bool rsort = 1;
  for (int i = 1; i < v.size(); ++i) {
    if (v[i - 1].second > v[i].second) {
      rsort = 0;
    }
  }
  if (rsort) {
    cnt12 = 1LL;
    for (int i = 1; i < v1.size(); ++i) {
      if (v[i] == v[i - 1]) {
        ++di;
      } else {
        cnt12 *= fact[di];
        cnt12 %= MOD;
        di = 1LL;
      }
    }
    cnt12 *= fact[di];
    cnt12 %= MOD;
  }
  long long ans = tot - cnt1 - cnt2 + cnt12;
  ans = (ans + 10 * MOD) % MOD;
  cout << ans << endl;
}
