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 maxn = 300000 + 10;
const int mod = 998244353;
long long fac[maxn], A[maxn];
long long solve(map<int, vector<int>> &cnt) {
  long long res = 1;
  for (auto &x : cnt) {
    res = ((res % mod) * (fac[x.second.size()] % mod)) % mod;
  }
  return res;
}
int main() {
  fac[0] = fac[1] = 1;
  for (int i = 2; i < maxn; i++)
    fac[i] = ((fac[i - 1] % mod) * (i % mod)) % mod;
  int N;
  cin >> N;
  map<int, vector<int>> cnt1, cnt2;
  for (int i = 0; i < N; i++) {
    int a, b;
    cin >> a >> b;
    cnt1[a].push_back(i);
    cnt2[b].push_back(i);
    A[i] = b;
  }
  int ant = 0;
  long long r2 = 1;
  for (auto &x : cnt1) {
    map<int, int> cnt3;
    for (auto &act : x.second) cnt3[A[act]]++;
    auto it = cnt3.begin();
    if (it->first < ant) {
      r2 = 0;
      break;
    }
    for (auto &act : cnt3) {
      r2 = ((fac[act.second] % mod) * (r2 % mod)) % mod;
      ant = max(ant, act.first);
    }
  }
  long long res = fac[N];
  res = ((res % mod) - (solve(cnt1))) % mod;
  res = ((res % mod) - (solve(cnt2))) % mod;
  res = ((res % mod) + (r2 % mod)) % mod;
  if (res < 0) res += mod;
  cout << res << '\n';
}
