#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 3e5 + 1;
const ll mod = 998244353LL;
int f[2][N];
ll fac[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  fac[0] = 1LL;
  for (int i = 1; i <= n; i += 1) fac[i] = (fac[i - 1] * 1LL * i) % mod;
  vector<pair<int, int>> v;
  for (int i = 1; i <= n; i += 1) {
    int x, y;
    cin >> x >> y;
    f[0][x] += 1;
    f[1][y] += 1;
    v.push_back({x, y});
  }
  ll ans = fac[n], p = 1LL, q = 1LL, r = 1LL;
  for (int i = 1; i <= n; i += 1) {
    p = (p * fac[f[0][i]]) % mod;
    q = (q * fac[f[1][i]]) % mod;
  }
  ans -= p;
  ans += mod;
  ans %= mod;
  ans -= q;
  ans += mod;
  ans %= mod;
  sort(v.begin(), v.end());
  int flag = 1;
  for (int i = 0; i < n; i += 1) {
    if (i && v[i].second < v[i - 1].second) {
      flag = 0;
      break;
    }
  }
  if (flag == 1) {
    int x = 1, y = 1;
    for (int i = 0; i < n; i += 1) {
      if (i && v[i].second == v[i - 1].second)
        x += 1;
      else {
        x = min(x, y);
        r = (r * fac[x]) % mod;
        x = 1;
        y = 1;
        continue;
      }
      if (i && v[i].first == v[i - 1].first)
        y += 1;
      else {
        x = min(x, y);
        r = (r * fac[x]) % mod;
        x = 1;
        y = 1;
        continue;
      }
    }
    x = min(x, y);
    r = (r * fac[x]) % mod;
    ans += r;
    ans %= mod;
  }
  cout << ans;
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 3e5 + 1;
const ll mod = 998244353LL;
int f[2][N];
ll fac[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  fac[0] = 1LL;
  for (int i = 1; i <= n; i += 1) fac[i] = (fac[i - 1] * 1LL * i) % mod;
  vector<pair<int, int>> v;
  for (int i = 1; i <= n; i += 1) {
    int x, y;
    cin >> x >> y;
    f[0][x] += 1;
    f[1][y] += 1;
    v.push_back({x, y});
  }
  ll ans = fac[n], p = 1LL, q = 1LL, r = 1LL;
  for (int i = 1; i <= n; i += 1) {
    p = (p * fac[f[0][i]]) % mod;
    q = (q * fac[f[1][i]]) % mod;
  }
  ans -= p;
  ans += mod;
  ans %= mod;
  ans -= q;
  ans += mod;
  ans %= mod;
  sort(v.begin(), v.end());
  int flag = 1;
  for (int i = 0; i < n; i += 1) {
    if (i && v[i].second < v[i - 1].second) {
      flag = 0;
      break;
    }
  }
  if (flag == 1) {
    int x = 1, y = 1;
    for (int i = 0; i < n; i += 1) {
      if (i && v[i].second == v[i - 1].second)
        x += 1;
      else {
        x = min(x, y);
        r = (r * fac[x]) % mod;
        x = 1;
        y = 1;
        continue;
      }
      if (i && v[i].first == v[i - 1].first)
        y += 1;
      else {
        x = min(x, y);
        r = (r * fac[x]) % mod;
        x = 1;
        y = 1;
        continue;
      }
    }
    x = min(x, y);
    r = (r * fac[x]) % mod;
    ans += r;
    ans %= mod;
  }
  cout << ans;
  return 0;
}

```