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 = 300005, md = 998244353;
int n, ans, sa, sb, sc;
struct node {
  int a, b;
} nm[N];
int fac[N];
bool cmp1(node a, node b) {
  if (a.a == b.a) return a.b < b.b;
  return a.a < b.a;
}
bool cmp2(node a, node b) { return a.b < b.b; }
template <class I>
inline void get(I &a) {
  char c = getchar();
  a = 0;
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; c = getchar()) a = a * 10 + c - '0';
}
int main() {
  get(n);
  for (int i = 1; i <= n; i++) {
    get(nm[i].a);
    get(nm[i].b);
  }
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = (long long)fac[i - 1] * i % md;
  sort(nm + 1, nm + 1 + n, cmp1);
  sa = sb = sc = 1;
  ans = fac[n];
  int la = 1;
  for (int i = 1; i <= n; i++)
    if (i != 1 && nm[i].a != nm[i - 1].a) {
      sa = (long long)sa * fac[i - la] % md;
      la = i;
    }
  sa = (long long)sa * fac[n - la + 1] % md;
  la = 1;
  for (int i = 1; i <= n; i++)
    if (!(nm[i].a >= nm[i - 1].a && nm[i].b >= nm[i - 1].b)) {
      sb = 0;
      break;
    }
  for (int i = 1; i <= n; i++)
    if (i != 1 && (nm[i].a != nm[i - 1].a || nm[i].b != nm[i - 1].b)) {
      sb = (long long)sb * fac[i - la] % md;
      la = i;
    }
  sb = (long long)sb * fac[n - la + 1] % md;
  sort(nm + 1, nm + 1 + n, cmp2);
  la = 1;
  for (int i = 1; i <= n; i++)
    if (i != 1 && nm[i].b != nm[i - 1].b) {
      sc = (long long)sc * fac[i - la] % md;
      la = i;
    }
  sc = (long long)sc * fac[n - la + 1] % md;
  ans = ((long long)ans - sa - sc + sb + 2 * md) % md;
  printf("%d\n", ans);
  return 0;
}
