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;
struct node {
  long long x;
  long long y;
} p[300005];
long long jc[300005] = {0};
inline long long rd() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * w;
}
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.y < b.y;
  else
    return a.x < b.x;
}
int main() {
  long long n;
  cin >> n;
  jc[0] = 1;
  for (int i = 1; i <= n; i++) {
    jc[i] = jc[i - 1] * i % 998244353;
  }
  for (int i = 0; i < n; i++) {
    cin >> p[i].x >> p[i].y;
  }
  sort(p, p + n, cmp1);
  long long ans1 = 1;
  for (int i = 1; i < n; i++) {
    long long t = 1;
    while (i < n && p[i].x == p[i - 1].x) {
      t++;
      i++;
    }
    ans1 = ((ans1 + 998244353) % 998244353 * jc[t] + 998244353) % 998244353;
  }
  sort(p, p + n, cmp2);
  long long ans2 = 1;
  for (int i = 1; i < n; i++) {
    long long t = 1;
    while (i < n && p[i].y == p[i - 1].y) {
      t++;
      i++;
    }
    ans2 = ((ans2 + 998244353) % 998244353 * jc[t] + 998244353) % 998244353;
  }
  sort(p, p + n, cmp3);
  long long ans3 = 1;
  for (int i = 1; i < n; i++) {
    if (p[i].y < p[i - 1].y) ans3 = 0;
  }
  for (int i = 1; i < n; i++) {
    long long t = 1;
    while (i < n && p[i].x == p[i - 1].x && p[i].y == p[i - 1].y) {
      t++;
      i++;
    }
    ans3 = ((ans3 + 998244353) % 998244353 * jc[t] + 998244353) % 998244353;
  }
  long long ans;
  ans = (jc[n] - ans1 - ans2 + ans3) % 998244353;
  ans += 998244353;
  ans %= 998244353;
  cout << ans;
  return 0;
}
