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;
struct S {
  int a, b;
  S() {}
  S(int _a, int _b) {
    a = _a;
    b = _b;
  }
  const bool operator<(const S &o) const { return a < o.a; }
};
string exm;
inline void exf(void) {
  cout << exm << "\n";
  exit(0);
}
template <typename T>
inline void showAll(vector<T> &v, string sep = "") {
  for (T &here : v) cout << here << sep;
}
template <typename T>
inline void showAll(T arr[], int st, int end, string sep = "") {
  for (int i = st; i <= end; i++) cout << arr[i] << sep;
}
template <typename T>
inline vector<int> int_seperation(T N, int d = 10) {
  vector<int> v;
  while (N) {
    v.push_back(N % d);
    N /= d;
  }
  reverse(v.begin(), v.end());
  return v;
}
const int SIZE = 300009;
const int MOD = 998244353;
pair<int, int> arr[SIZE];
map<int, int> a, b;
map<pair<int, int>, int> c;
long long fact[SIZE];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &arr[i].first, &arr[i].second);
    a[arr[i].first]++;
    b[arr[i].second]++;
    c[{arr[i].first, arr[i].second}]++;
  }
  fact[0] = 1;
  for (int i = 1; i <= n; i++) {
    fact[i] = (fact[i - 1] * i) % MOD;
  }
  sort(arr + 1, arr + 1 + n);
  long long ans = fact[n];
  long long sub_a, sub_b, sub_c;
  sub_a = sub_b = sub_c = 1;
  for (auto here : a) {
    int sec = here.second;
    sub_a = sub_a * fact[sec] % MOD;
  }
  for (auto here : b) {
    int sec = here.second;
    sub_b = sub_b * fact[sec] % MOD;
  }
  for (auto here : c) {
    int sec = here.second;
    sub_c = sub_c * fact[sec] % MOD;
  }
  for (int i = 1; i <= n - 1; i++) {
    if (arr[i].second > arr[i + 1].second) sub_c = 0;
  }
  ans = (ans - sub_a - sub_b + sub_c);
  while (ans < 0) {
    ans += MOD;
  }
  ans %= MOD;
  printf("%lld\n", ans);
  return 0;
}
