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 + 5;
const int mod = 998244353;
int n;
pair<int, int> p[N];
long long val[N];
int c[N];
int comp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
void init() {
  val[1] = 1;
  for (int i = 2; i <= n; i++) val[i] = (val[i - 1] * i) % mod;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &p[i].first, &p[i].second);
  }
  init();
  sort(p + 1, p + n + 1, comp);
  int s = 1;
  long long ans1 = 1;
  int fl = 0;
  int ma = 0;
  while (s <= n) {
    int ss = s;
    int tmp1 = 0, tmp2 = 1000000;
    while (ss <= n && p[ss].second == p[s].second)
      tmp1 = max(tmp1, p[ss].first), tmp2 = min(tmp2, p[ss].first), ss++;
    ans1 = (ans1 * val[ss - s]) % mod;
    if (tmp2 < ma) fl = 1;
    ma = max(ma, tmp1);
    s = ss;
  }
  sort(p + 1, p + n + 1);
  s = 1;
  ma = 0;
  long long ans2 = 1;
  while (s <= n) {
    int ss = s;
    int tmp1 = 0, tmp2 = 1000000;
    while (ss <= n && p[ss].first == p[s].first)
      tmp1 = max(tmp1, p[ss].second), tmp2 = min(tmp2, p[ss].second), ss++;
    ans2 = (ans2 * val[ss - s]) % mod;
    if (tmp2 < ma) fl = 1;
    ma = max(ma, tmp1);
    s = ss;
  }
  if (fl) {
    printf("%I64d\n", (val[n] - ans1 - ans2 + 2 * mod) % mod);
  } else {
    int s = 1;
    int cnt = 0;
    long long ans3 = 1;
    while (s <= n) {
      int ss = s;
      cnt = 0;
      while (ss <= n && p[ss].first == p[s].first)
        c[cnt++] = p[ss].second, ss++;
      sort(c, c + cnt);
      int t = 0;
      while (t < cnt) {
        int tt = t;
        while (tt < cnt && c[tt] == c[t]) tt++;
        ans3 = (ans3 * val[tt - t]) % mod;
        t = tt;
      }
      s = ss;
    }
    printf("%I64d\n", (val[n] - ans1 - ans2 + ans3 + 2 * mod) % mod);
  }
  return 0;
}
