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;
void read(int &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -f;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  x *= f;
}
void print(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (!x) return;
  print(x / 10), putchar(x % 10 + 48);
}
void write(int x) {
  if (!x)
    putchar('0');
  else
    print(x);
  putchar('\n');
}
const int maxn = 1e6 + 10;
const int inf = 1e9;
const double eps = 1e-8;
const int mod = 998244353;
pair<int, int> a[maxn];
int n, fac[maxn], ans;
int cmp(pair<int, int> x, pair<int, int> y) { return x.second < y.second; }
int main() {
  read(n);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * i * fac[i - 1] % mod;
  for (int i = 1; i <= n; i++) read(a[i].first), read(a[i].second);
  int s = 1;
  sort(a + 1, a + n + 1);
  int tmp = 1;
  for (int i = 2; i <= n + 1; i++)
    if (i <= n && a[i].second < a[i - 1].second) {
      tmp = 0;
      break;
    } else if (a[i] == a[i - 1])
      s++;
    else
      tmp = 1ll * tmp * fac[s] % mod, s = 1;
  ans = (mod - tmp) % mod;
  tmp = 1;
  s = 1;
  for (int i = 2; i <= n + 1; i++)
    if (a[i].first == a[i - 1].first)
      s++;
    else
      tmp = 1ll * tmp * fac[s] % mod, s = 1;
  sort(a + 1, a + n + 1, cmp);
  ans = (ans + tmp) % mod;
  tmp = s = 1;
  for (int i = 2; i <= n + 1; i++)
    if (a[i].second == a[i - 1].second)
      s++;
    else
      tmp = 1ll * tmp * fac[s] % mod, s = 1;
  ans = (ans + tmp) % mod;
  write((fac[n] - ans + mod) % mod);
  return 0;
}
