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;
double EPS = 1e-9;
int INF = 1e9 + 7;
;
long long INFLL = 1e17;
double pi = acos(-1);
int dirx[8] = {-1, 0, 0, 1, -1, -1, 1, 1};
int diry[8] = {0, 1, -1, 0, -1, 1, -1, 1};
const int MAXN = 3e5 + 5;
const int MOD = 998244353;
int n;
long long fac[MAXN];
long long solve(vector<pair<int, int> > A) {
  sort(A.begin(), A.end());
  int last = -1;
  long long ans = 1;
  for (int k = (int)0; k < (int)A.size(); k++) {
    if (k == (int)A.size() - 1 || A[k].first != A[k + 1].first) {
      ans *= fac[k - last];
      ans %= MOD;
      last = k;
    }
  }
  return ans;
}
long long solve2(vector<pair<int, int> > A) {
  sort(A.begin(), A.end());
  int last = -1;
  long long ans = 1;
  for (int k = (int)1; k < (int)A.size(); k++)
    if (!(A[k - 1].first <= A[k].first && A[k - 1].second <= A[k].second))
      return 0;
  for (int k = (int)0; k < (int)A.size(); k++) {
    if (k == (int)A.size() - 1 || A[k] != A[k + 1]) {
      ans *= fac[k - last];
      ans %= MOD;
      last = k;
    }
  }
  return ans;
}
vector<pair<int, int> > A;
int main() {
  fac[0] = 1;
  for (int k = (int)1; k < (int)MAXN; k++) fac[k] = k * fac[k - 1] % MOD;
  cin >> n;
  A.resize(n);
  for (int k = (int)0; k < (int)n; k++) cin >> A[k].first >> A[k].second;
  long long ans = fac[n] - solve(A);
  for (int k = (int)0; k < (int)n; k++) swap(A[k].first, A[k].second);
  ans -= solve(A);
  ans += solve2(A);
  ans %= MOD;
  if (ans < 0) ans += MOD;
  cout << ans << endl;
}
