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 mod = 998244353;
const int p = 3e5 + 5;
unordered_map<int, int> m1, m2;
unordered_map<long long, int> m3;
int a[p], b[p];
int fact[p];
long long nap(long long a, long long b) { return a * p + b; }
inline int mul(int a, int b) { return (long long)a * (long long)b % mod; }
inline int sum(int a, int b) {
  if (a + b >= mod) {
    return a + b - mod;
  }
  if (a + b < 0) {
    return a + b + mod;
  }
  return a + b;
}
void precompute() {
  fact[0] = 1;
  for (int i = 1; i < p; i++) {
    fact[i] = mul(fact[i - 1], i);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  precompute();
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
    m1[a[i]]++;
    m2[b[i]]++;
    m3[nap(a[i], b[i])]++;
  }
  sort(a, a + n);
  sort(b, b + n);
  int sol = fact[n];
  int br1 = 1, br2 = 1;
  for (int i = 0; i < p; i++) {
    if (m1[i]) {
      br1 = mul(br1, fact[m1[i]]);
    }
    if (m2[i]) {
      br2 = mul(br2, fact[m2[i]]);
    }
  }
  sol = sum(sol, -br1);
  sol = sum(sol, -br2);
  br1 = 1;
  int br = 0;
  for (int i = 0; i < n; i++) {
    if (i > 0 && a[i - 1] == a[i] && b[i - 1] == b[i]) {
      continue;
    }
    if (!m3[nap(a[i], b[i])]) {
      br1 = 0;
      break;
    }
    br1 = mul(br1, fact[m3[nap(a[i], b[i])]]);
    br += m3[nap(a[i], b[i])];
  }
  if (br != n) {
    br1 = 0;
  }
  sol = sum(sol, br1);
  cout << sol << '\n';
  return 0;
}
