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 long long mod = 998244353;
int n;
long long fac[300005];
long long ans = 1;
long long ans1 = 1;
long long ans2 = 1;
long long ans3 = 1;
struct Node {
  int x, y;
} s[300005];
bool cmp1(Node a, Node b) { return a.x < b.x; }
bool cmp2(Node a, Node b) { return a.y < b.y; }
bool cmp3(Node a, Node b) {
  if (a.x != b.x)
    return a.x < b.x;
  else
    return a.y < b.y;
}
int main() {
  int i, j, k, p;
  bool flag = 1;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d%d", &s[i].x, &s[i].y);
  for (i = 1; i <= n; i++) {
    ans = (ans * i) % mod;
    fac[i] = ans;
  }
  sort(s + 1, s + n + 1, cmp1);
  for (i = 1; i <= n; i++) {
    j = i;
    while (j + 1 <= n && s[j + 1].x == s[i].x) j++;
    if (j - i + 1 == 1)
      continue;
    else
      ans1 = (ans1 * fac[j - i + 1]) % mod;
    i = j;
  }
  sort(s + 1, s + n + 1, cmp2);
  for (i = 1; i <= n; i++) {
    j = i;
    while (j + 1 <= n && s[j + 1].y == s[i].y) j++;
    if (j - i + 1 == 1)
      continue;
    else
      ans2 = (ans2 * fac[j - i + 1]) % mod;
    i = j;
  }
  sort(s + 1, s + n + 1, cmp3);
  for (i = 2; i <= n; i++)
    if (s[i].y < s[i - 1].y) flag = 0;
  if (!flag)
    ans3 = 0;
  else {
    for (i = 1; i <= n; i++) {
      j = i;
      while (j + 1 <= n && s[j + 1].x == s[i].x) j++;
      if (j - i + 1 == 1)
        continue;
      else {
        for (k = i; k <= j; k++) {
          p = k;
          while (p + 1 <= j && s[p + 1].y == s[k].y) p++;
          ans3 = (ans3 * fac[p - k + 1]) % mod;
          k = p;
        }
      }
      i = j;
    }
  }
  ans = (ans - ans1 - ans2 + ans3 + mod * 5) % mod;
  printf("%lld", ans);
}
