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 = 3e5 + 15;
const int mod = 998244353;
int add(int x, int y) {
  int ret = (x + y) % mod;
  ret = (ret + mod) % mod;
  return ret;
}
int mul(int x, int y) {
  long long ret = x;
  ret *= y;
  ret %= mod;
  ret = (ret + mod) % mod;
  return (int)ret;
}
pair<int, int> in[N];
int per[N];
int get(int n) {
  int tmp = 1;
  for (int i = 0; i < n; ++i) {
    swap(in[i].first, in[i].second);
  }
  sort(in, in + n);
  for (int i = 0; i < n;) {
    int j;
    for (j = i; j < n; ++j) {
      if (in[j].first != in[i].first) {
        break;
      }
    }
    tmp = mul(tmp, per[j - i]);
    i = j;
  }
  return tmp;
}
void solve() {
  int n;
  per[0] = 1;
  scanf("%d", &n);
  for (int i = 1; i < N; ++i) {
    per[i] = mul(per[i - 1], i);
  }
  for (int i = 0; i < n; ++i) {
    scanf("%d %d", &in[i].first, &in[i].second);
  }
  int ret = per[n];
  int tmp = 1;
  sort(in, in + n);
  bool g = true;
  int maxi = in[0].second;
  for (int i = 1; i < n; ++i) {
    if (maxi > in[i].second) {
      g = false;
      break;
    }
    maxi = max(maxi, in[i].second);
  }
  int rr = 0;
  if (g) {
    int l = 0;
    int d = 1;
    while (l < n) {
      int r = l + 1;
      while (r < n && in[l].first == in[r].first) ++r;
      map<int, int> m;
      for (int i = l; i < r; ++i) ++m[in[i].second];
      for (auto p : m) d = mul(d, per[p.second]);
      l = r;
    }
    rr = -d;
  }
  int v = add(get(n), get(n));
  rr = add(v, rr);
  ret = add(ret, -rr);
  printf("%d\n", ret);
}
int main(void) { solve(); }
