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 INF = 0x3f3f3f3f;
mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
template <typename... Args>
void write(Args... args) {
  ((cout << args << " "), ...);
  cout << '\n';
}
struct myhash {
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    x ^= FIXED_RANDOM;
    return x ^ (x >> 16);
  }
};
constexpr int mod = 998244353;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  unordered_map<int, int, myhash> c1, c2;
  vector<pair<int, int>> a(n);
  for (auto& [x, y] : a) {
    cin >> x >> y;
    c1[x]++;
    c2[y]++;
  }
  sort((a).begin(), (a).end());
  vector<long long> fac(n + 1);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) {
    fac[i] = fac[i - 1] * i % mod;
  }
  long long cnt1 = 1, cnt2 = 1, ans = fac[n];
  for (auto [_, y] : c1) cnt1 = cnt1 * fac[y] % mod;
  (ans -= cnt1 - mod) %= mod;
  for (auto [_, y] : c2) cnt2 = cnt2 * fac[y] % mod;
  (ans -= cnt2 - mod) %= mod;
  bool issorted = 1;
  for (int i = 1; i < n; i++) {
    if (a[i].second < a[i - 1].second) {
      issorted = 0;
      break;
    }
  }
  if (issorted) {
    cnt1 = 1;
    for (int i = 0, j = 0; i < n; i = j) {
      while (j < n && a[j] == a[i]) j++;
      (cnt1 *= fac[j - i]) %= mod;
    }
    (ans += cnt1) %= mod;
  }
  cout << ans;
  return 0;
}
