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;
bool compare(const pair<int, int>& i, const pair<int, int>& j) {
  if (i.first != j.first) {
    return i.first < j.first;
  } else {
    return i.second < j.second;
  }
}
int main() {
  int n;
  const long long p = 998244353;
  cin >> n;
  pair<int, int> pairs[n];
  pair<int, int> tmp;
  for (int i = 0; i < n; i++) {
    cin >> tmp.first >> tmp.second;
    pairs[i] = tmp;
  }
  sort(pairs, pairs + n, compare);
  long long ans = 0;
  vector<int> sizes;
  vector<int> sizes2;
  vector<int> intersectSizes;
  long long factorials[n + 1];
  factorials[0] = 1;
  for (int i = 1; i <= n; i++) {
    factorials[i] = (factorials[i - 1] * i) % p;
  }
  int curr = 1;
  int currIntersect = 1;
  bool isThereIntersect = true;
  for (int i = 1; i < n; i++) {
    if (pairs[i].first == pairs[i - 1].first) {
      curr++;
    } else if (curr > 1) {
      sizes.push_back(curr);
      curr = 1;
    }
    if (pairs[i].second < pairs[i - 1].second) {
      isThereIntersect = false;
    }
    if (pairs[i].first == pairs[i - 1].first &&
        pairs[i].second == pairs[i - 1].second) {
      currIntersect++;
    } else if (currIntersect > 1) {
      intersectSizes.push_back(currIntersect);
      currIntersect = 1;
    }
  }
  sizes.push_back(curr);
  intersectSizes.push_back(currIntersect);
  int temp;
  for (int i = 0; i < n; i++) {
    temp = pairs[i].first;
    pairs[i].first = pairs[i].second;
    pairs[i].second = temp;
  }
  sort(pairs, pairs + n, compare);
  curr = 1;
  for (int i = 1; i < n; i++) {
    if (pairs[i].first == pairs[i - 1].first) {
      curr++;
    } else if (curr > 1) {
      sizes2.push_back(curr);
      curr = 1;
    }
  }
  sizes2.push_back(curr);
  long long first = 1;
  long long second = 1;
  long long intersect = 1;
  for (int i = 0; i < sizes.size(); i++) {
    first = (first * factorials[sizes[i]]) % p;
  }
  for (int i = 0; i < sizes2.size(); i++) {
    second = (second * factorials[sizes2[i]]) % p;
  }
  for (int i = 0; i < intersectSizes.size(); i++) {
    intersect = (intersect * factorials[intersectSizes[i]]) % p;
  }
  if (isThereIntersect == false) {
    intersect = 0;
  }
  ans = ((factorials[n] - ((first + second) % p - intersect) % p) % p + p) % p;
  cout << ans;
}
