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 Pair {
  int f, s;
};
const int maxn = 3e5 + 20;
const int MOD = 998244353;
long long f[maxn];
Pair a[maxn];
int F(long long x) {
  if (x == 1) return f[x] = 1;
  return f[x] = (x * F(x - 1)) % MOD;
}
bool cmp(Pair a, Pair b) { return a.f < b.f; }
bool cmp2(Pair a, Pair b) { return (a.f != b.f) ? a.f < b.f : a.s < b.s; }
int main() {
  int n;
  cin >> n;
  long long ans = F(n);
  for (int i = 0; i < n; i++) cin >> a[i].f >> a[i].s;
  for (int k = 0; k < 2; k++) {
    sort(a, a + n, cmp);
    long long tp = 1;
    int top = a[0].f;
    int cnt = 1;
    for (int j = 1; j < n; j++) {
      if (a[j].f != top) {
        tp = (tp * f[cnt]) % MOD;
        cnt = 1;
        top = a[j].f;
      } else
        cnt++;
    }
    tp = (tp * f[cnt]) % MOD;
    ans = (ans - tp) % MOD;
    for (int i = 0; i < n; i++) swap(a[i].f, a[i].s);
  }
  sort(a, a + n, cmp2);
  long long tp = 1;
  Pair top = a[0];
  int cnt = 1;
  for (int j = 1; j < n; j++) {
    if (a[j].s < top.s) tp = 0;
    if (a[j].f != top.f || a[j].s != top.s) {
      tp = (tp * f[cnt]) % MOD;
      cnt = 1;
      top = a[j];
    } else
      cnt++;
  }
  tp = (tp * f[cnt]) % MOD;
  ans = (ans + tp) % MOD;
  if (ans < 0) ans += MOD;
  cout << ans << endl;
}
