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;
using ll = long long;
const int N = 3e5 + 5;
const int MOD = 998244353;
int n, a[N][2];
int h1[N], h2[N], idx[N], idx2[N];
ll fm[N];
bool cmp(int i, int j) {
  if (a[i][0] < a[j][0]) return true;
  if (a[i][0] > a[j][0]) return false;
  if (a[i][1] < a[j][1]) return true;
  return false;
}
bool cmp2(int i, int j) {
  if (a[idx[i]][1] < a[idx[j]][1]) return true;
  if (a[idx[i]][1] > a[idx[j]][1]) return false;
  if (a[idx[i]][0] < a[idx[j]][0]) return true;
  return false;
}
long long pm(int i, int r) {
  for (int l = i; l <= r; l++) idx2[l] = l;
  sort(idx2 + i, idx2 + r + 1, cmp2);
  ll res = 1;
  int cnt = 0, last = -1;
  for (int l = i; l <= r; l++) {
    if (a[idx[idx2[l]]][1] != last) {
      res = (res * fm[cnt]) % MOD;
      cnt = 1;
      last = a[idx[idx2[l]]][1];
    } else
      cnt++;
  }
  res = (res * fm[cnt]) % MOD;
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  fm[0] = fm[1] = 1;
  for (int i = 1; i <= N - 2; i++) fm[i + 1] = (fm[i] * (i + 1)) % MOD;
  cin >> n;
  for (int c = 0; c < n; c++) {
    cin >> a[c][0] >> a[c][1];
    h1[a[c][0]]++;
    h2[a[c][1]]++;
    idx[c] = c;
  }
  ll ans = fm[n];
  ll f1 = 1, f2 = 1;
  for (int i = 0; i <= n; i++) {
    f1 = (f1 * fm[h1[i]]) % MOD;
    f2 = (f2 * fm[h2[i]]) % MOD;
  }
  ans = (ans + MOD - f1 + MOD - f2) % MOD;
  sort(idx, idx + n, cmp);
  ll fb = 1;
  for (int i = 0; i < n - 1; i++)
    if (a[idx[i]][1] > a[idx[i + 1]][1]) fb = 0;
  for (int i = 0; i < n; i++) {
    int r = i;
    for (r = i; r < n; r++)
      if (a[idx[r]][0] != a[idx[i]][0]) break;
    r--;
    fb = (fb * pm(i, r)) % MOD;
    i = r;
  }
  ans = (ans + fb) % MOD;
  cout << ans << endl;
  return 0;
}
