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 mod = 998244353;
int n, i, j, m;
long long ans, p[300005], s;
struct point {
  int x;
  int y;
} a[300005];
bool cmp(point a, point b) {
  if (a.x == b.x) return a.y < b.y;
  return a.x < b.x;
}
bool cmp2(point a, point b) { return a.y < b.y; }
int main() {
  scanf("%d", &n);
  p[0] = 1;
  for (i = 1; i <= n; i++) {
    scanf("%d %d", &a[i].x, &a[i].y);
    p[i] = p[i - 1] * i % mod;
  }
  ans = p[n];
  sort(a + 1, a + 1 + n, cmp);
  int flag = 0;
  for (i = 2; i <= n; i++)
    if (!(a[i].x >= a[i - 1].x && a[i].y >= a[i - 1].y)) flag = 1;
  if (!flag) {
    s = 1;
    for (i = 1; i <= n;) {
      for (j = i; j <= n && a[i].x == a[j].x && a[i].y == a[j].y; j++)
        ;
      s = s * p[j - i] % mod;
      i = j;
    }
    ans += s;
  }
  s = 1;
  for (i = 1; i <= n;) {
    for (j = i; j <= n && a[i].x == a[j].x; j++)
      ;
    s = s * p[j - i] % mod;
    i = j;
  }
  ans -= s;
  sort(a + 1, a + 1 + n, cmp2);
  s = 1;
  for (i = 1; i <= n;) {
    for (j = i; j <= n && a[i].y == a[j].y; j++)
      ;
    s = s * p[j - i] % mod;
    i = j;
  }
  ans -= s;
  printf("%lld\n", ((ans % mod) + mod) % mod);
}
