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;
using ll = long long;
using pii = pair<int, int>;
const int MOD = 998244353;
const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};
const char dir[] = {'R', 'L', 'D', 'U'};
ll add(ll a, ll b) {
  a += b;
  if (a >= MOD) {
    a -= MOD;
  }
  return a;
}
int sub(int a, int b) {
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
int mult(int a, int b) { return ((ll)a * b) % MOD; }
void setIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
int main() {
  setIO();
  int n;
  cin >> n;
  vector<pii> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i].first >> a[i].second;
  }
  int fact[n + 1];
  fact[1] = 1;
  for (int i = 2; i <= n; ++i) {
    fact[i] = mult(fact[i - 1], i);
  }
  int cnt[n + 1];
  fill(cnt, cnt + n + 1, 0);
  vector<int> grps;
  for (int i = 0; i < n; ++i) {
    ++cnt[a[i].first];
  }
  for (int i = 1; i <= n; ++i) {
    if (cnt[i]) grps.emplace_back(cnt[i]);
  }
  int ways1 = 1;
  for (int i : grps) {
    ways1 = mult(ways1, fact[i]);
  }
  int cnt2[n + 1];
  fill(cnt2, cnt2 + n + 1, 0);
  vector<int> grps2;
  for (int i = 0; i < n; ++i) {
    ++cnt2[a[i].second];
  }
  for (int i = 1; i <= n; ++i) {
    if (cnt2[i]) grps2.emplace_back(cnt2[i]);
  }
  int ways2 = 1;
  for (int i : grps2) {
    ways2 = mult(ways2, fact[i]);
  }
  int waysall = 1;
  map<pii, int> both;
  for (int i = 0; i < n; ++i) {
    ++both[a[i]];
  }
  vector<int> grpsboth;
  for (auto it = both.begin(); it != both.end(); ++it) {
    grpsboth.emplace_back((*it).second);
  }
  for (int i : grpsboth) {
    waysall = mult(waysall, fact[i]);
  }
  sort((a).begin(), (a).end());
  bool can = true;
  int mxx = 0, mxy = 0;
  for (int i = 0; i < n; ++i) {
    if (a[i].first < mxx || a[i].second < mxy) can = false;
    mxx = max(mxx, a[i].first);
    mxy = max(mxy, a[i].second);
  }
  if (!can) waysall = 0;
  int cant = sub(add(ways1, ways2), waysall);
  cout << sub(fact[n], cant) << "\n";
}
