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;
inline int read() {
  char ch = getchar();
  int x = 0, f = 1;
  while (ch < '0' || ch > '9') f = ch == '-' ? -1 : 1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
const int MAXN = 300010;
const int mod = 998244353;
long long fac[MAXN];
struct node {
  int a, b;
  node() { a = b = 0; }
  node(int a, int b) { this->a = a, this->b = b; }
} p[MAXN];
bool cmp1(const node &a, const node &b) { return a.a < b.a; }
bool cmp2(const node &a, const node &b) { return a.b < b.b; }
bool cmp3(const node &a, const node &b) {
  if (a.a == b.a)
    return a.b < b.b;
  else
    return a.a < b.a;
}
int main(int argc, char *argv[]) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n = read();
  fac[0] = fac[1] = 1;
  for (int i = (2); i <= (n); i += (1)) fac[i] = (fac[i - 1] * i) % mod;
  for (int i = (1); i <= (n); i += (1)) p[i].a = read(), p[i].b = read();
  long long res1 = 1, res2 = 1, res3 = 1;
  int l, r;
  sort(p + 1, p + n + 1, cmp1);
  l = 1, r = 1;
  while (l <= n) {
    while (r <= n && p[l].a == p[r].a) r++;
    res1 = (res1 * fac[r - l]) % mod;
    l = r;
  }
  sort(p + 1, p + n + 1, cmp2);
  l = 1, r = 1;
  while (l <= n) {
    while (r <= n && p[l].b == p[r].b) r++;
    res2 = (res2 * fac[r - l]) % mod;
    l = r;
  }
  sort(p + 1, p + n + 1, cmp3);
  for (int i = (2); i <= (n); i += (1))
    if (p[i].b < p[i - 1].b) res3 = 0;
  if (res3 != 0) {
    l = 1, r = 1;
    while (l <= n) {
      while (r <= n && p[l].a == p[r].a && p[l].b == p[r].b) r++;
      if (res3 == 0) break;
      res3 = (res3 * fac[r - l]) % mod;
      l = r;
    }
  }
  long long res = (fac[n] - res1 - res2 + res3) % mod;
  if (res < 0) res += mod;
  printf("%lld\n", res);
  return 0;
}
