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 long long int INF = 9223372036854775807;
const long long int NINF = -100000000000000000;
const long long int mod = 998244353;
long long int power(long long int x, long long int y) {
  if (y == 0) return 1;
  long long int p = power(x, y / 2) % mod;
  p = (p * p) % mod;
  return (y % 2 == 0) ? p : (x * p) % mod;
}
long long int divide(long long int a, long long int b) {
  if (a % b == 0) {
    return a / b;
  } else
    return a / b + 1;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int fact[300001];
void init() {
  fact[0] = 1;
  fact[1] = 1;
  for (int i = 2; i <= 300000; i++) {
    fact[i] = (fact[i - 1] * i) % mod;
  }
}
bool eq(pair<int, int> a, pair<int, int> b) {
  if (a.first == b.first && a.second == b.second) return true;
  return false;
}
bool eq1(pair<int, int> a, pair<int, int> b) {
  if (a.first == b.second && a.second == b.first) return true;
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  init();
  int n;
  cin >> n;
  map<int, int> m;
  map<int, int> m1;
  map<pair<int, int>, int> m2;
  vector<pair<int, int> > a(n);
  vector<pair<int, int> > b(n);
  for (int i = 0; i < n; i++) {
    int ar, br;
    cin >> ar >> br;
    m1[br]++;
    m[ar]++;
    a[i] = make_pair(ar, br);
    b[i] = make_pair(br, ar);
    m2[a[i]]++;
  }
  sort(a.begin(), a.end());
  sort(b.begin(), b.end());
  int flag = 0;
  for (int i = 0; i < n; i++) {
    if (!eq1(a[i], b[i])) {
      flag = 1;
      break;
    }
  }
  long long int c = 1;
  if (flag == 0) {
    for (auto i : m2) {
      c = (c * fact[i.second]) % mod;
    }
  }
  if (flag == 1) {
    c = 0;
  }
  long long int c1 = 1, c2 = 1;
  for (auto i : m) {
    c1 = (c1 * fact[i.second]) % mod;
  }
  for (auto i : m1) {
    c2 = (c2 * fact[i.second]) % mod;
  }
  long long int out = fact[n];
  cout << (out + c - c1 - c2 + 2 * mod) % mod;
}
