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 = 3e5 + 10;
const int inf = 1e9 + 10;
const long long mo = 998244353;
struct node {
  int a, b;
} P[N];
bool cmp3(node x, node y) {
  if (x.a == y.a) return x.b < y.b;
  return x.a < y.a;
}
int mp1[N], mp2[N];
long long f[N];
int main() {
  int n;
  scanf("%d", &n);
  long long ans = 1;
  f[0] = 1;
  for (int i = 1; i <= n; i++) {
    f[i] = (f[i - 1] * i) % mo;
    scanf("%d %d", &P[i].a, &P[i].b);
    mp1[P[i].a]++;
    mp2[P[i].b]++;
    ans = (ans * i) % mo;
  }
  if (n == 1) {
    puts("0");
    return 0;
  }
  long long tmp1 = 1;
  long long tmp2 = 1;
  for (int i = 1; i <= n; i++) {
    tmp1 = (tmp1 * f[mp1[i]]) % mo;
    tmp2 = (tmp2 * f[mp2[i]]) % mo;
  }
  ans = (ans + mo - tmp1) % mo;
  ans = (ans + mo - tmp2) % mo;
  sort(P + 1, P + n + 1, cmp3);
  bool flag = true;
  long long tmp = 1;
  for (int i = 2; i <= n; i++) {
    if (P[i].b < P[i - 1].b) {
      flag = false;
      break;
    }
  }
  if (flag) {
    long long ans3 = mo - 1;
    node xx = node{-1, -1};
    long long cnt = 1;
    for (int i = 1; i <= n; i++) {
      if (P[i].a == xx.a && P[i].b == xx.b)
        cnt++;
      else {
        ans3 = (ans3 * f[cnt]) % mo;
        xx = P[i];
        cnt = 1;
      }
    }
    ans3 = (ans3 * f[cnt]) % mo;
    ans = (ans + mo - ans3) % mo;
  }
  printf("%I64d\n", ans);
  return 0;
}
