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 = 2e6 + 100;
const long long mod = 998244353;
template <class nmsl>
inline void read(nmsl &x) {
  x = 0;
  char ch = getchar(), w = 0;
  while (!isdigit(ch)) w = (ch == '-'), ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  x = w ? -x : x;
}
long long ans;
int n;
struct abc {
  int a, b;
} qwq[N];
bool cmp1(const abc &a, const abc &b) {
  if (a.a != b.a) return a.a < b.a;
  return a.b < b.b;
}
bool cmp2(const abc &a, const abc &b) { return a.b < b.b; }
long long jc[N];
int main() {
  read(n);
  jc[0] = 1;
  for (register int i = 1; i <= n; i++) jc[i] = jc[i - 1] * i % mod;
  ans = jc[n];
  for (register int i = 1; i <= n; i++) read(qwq[i].a), read(qwq[i].b);
  sort(qwq + 1, qwq + 1 + n, cmp2);
  long long tmp = 1, cnt = 1;
  for (register int i = 1; i <= n; i++) {
    if (qwq[i].b != qwq[i - 1].b)
      tmp = tmp * jc[cnt] % mod, cnt = 1;
    else
      cnt++;
  }
  tmp = tmp * jc[cnt] % mod;
  ans = ((ans - tmp) % mod + mod) % mod;
  sort(qwq + 1, qwq + 1 + n, cmp1);
  tmp = cnt = 1;
  for (register int i = 1; i <= n; i++) {
    if (qwq[i].a != qwq[i - 1].a)
      tmp = tmp * jc[cnt] % mod, cnt = 1;
    else
      cnt++;
  }
  tmp = tmp * jc[cnt] % mod;
  ans = ((ans - tmp) % mod + mod) % mod;
  for (register int i = 2; i <= n; i++) {
    if (qwq[i].b < qwq[i - 1].b) {
      cout << ans;
      return 0;
    }
  }
  tmp = cnt = 1;
  for (register int i = 1; i <= n; i++) {
    if (qwq[i].a != qwq[i - 1].a || qwq[i].b != qwq[i - 1].b)
      tmp = tmp * jc[cnt] % mod, cnt = 1;
    else
      cnt++;
  }
  tmp = tmp * jc[cnt] % mod;
  ans = ((ans + tmp) % mod + mod) % mod;
  cout << ans;
  return 0;
}
