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 N = int(3e5) + 99;
const int MOD = 998244353;
int mul(int a, int b) { return (a * 1LL * b) % MOD; }
int sum(int a, int b) { return (a + b) % MOD; }
int n;
pair<int, int> a[N];
int f[N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d%d", &a[i].first, &a[i].second);
  f[0] = 1;
  for (int i = 1; i < N; ++i) f[i] = mul(i, f[i - 1]);
  int res = f[n];
  for (int c = 0; c < 2; ++c) {
    int d = 1;
    sort(a, a + n);
    int l = 0;
    while (l < n) {
      int r = l + 1;
      while (r < n && a[l].first == a[r].first) ++r;
      d = mul(d, f[r - l]);
      l = r;
    }
    res = sum(res, MOD - d);
    for (int i = 0; i < n; ++i) swap(a[i].first, a[i].second);
  }
  sort(a, a + n);
  int l = 0;
  int d = 1;
  while (l < n) {
    int r = l + 1;
    while (r < n && a[l].first == a[r].first) ++r;
    map<int, int> m;
    for (int i = l; i < r; ++i) ++m[a[i].second];
    for (auto p : m) d = mul(d, f[p.second]);
    l = r;
  }
  for (int i = 1; i < n; ++i)
    if (a[i - 1].second > a[i].second) d = 0;
  res = sum(res, d);
  printf("%d\n", res);
  return 0;
}
