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 std::pair;
long long n, fac[300005], inv[300005], ans, inter = 1, l = 1, r = 1;
pair<long long, long long> a[300005];
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  return a.second < b.second;
}
int main() {
  scanf("%d", &n);
  fac[0] = fac[1] = inv[0] = inv[1] = 1;
  for (int i = 2; i <= n; ++i)
    fac[i] = (long long)fac[i - 1] * i % 998244353,
    inv[i] = (998244353 -
              (long long)inv[998244353 % i] * (998244353 / i) % 998244353);
  for (int i = 1; i <= n; ++i)
    inv[i] = (long long)inv[i - 1] * inv[i] % 998244353;
  for (int i = 1; i <= n; ++i) scanf("%d%d", &a[i].first, &a[i].second);
  std::sort(a + 1, a + n + 1);
  for (int i = 2; i <= n; ++i)
    if (a[i - 1].second > a[i].second) inter = 0;
  if (inter) {
    for (int i = 1, rr; i <= n; i = rr + 1) {
      rr = i;
      while (rr < n && a[rr + 1].second == a[i].second &&
             a[rr + 1].first == a[i].first)
        ++rr;
      inter = (long long)inter * fac[rr - i + 1] % 998244353;
    }
  }
  for (int i = 1, rr; i <= n; i = rr + 1) {
    rr = i;
    while (rr < n && a[rr + 1].first == a[i].first) ++rr;
    l = (long long)l * fac[rr - i + 1] % 998244353;
  }
  std::sort(a + 1, a + n + 1, cmp);
  for (int i = 1, rr; i <= n; i = rr + 1) {
    rr = i;
    while (rr < n && a[rr + 1].second == a[i].second) ++rr;
    r = (long long)r * fac[rr - i + 1] % 998244353;
  }
  printf("%d\n",
         (((long long)fac[n] - (long long)l - (long long)r + (long long)inter) %
              998244353 +
          (long long)998244353) %
             998244353);
  return 0;
}
