#include <bits/stdc++.h>
using namespace std;
int n;
long long ans1 = 1, ans2 = 1, ans3 = 1;
long long fact[400400];
int a[400400], b[400400];
pair<int, int> c[400400];
const long long mod = 998244353;
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i] >> b[i];
    c[i] = make_pair(a[i], b[i]);
  }
  fact[0] = 1;
  for (int i = 1; i <= n; ++i) fact[i] = (fact[i - 1] * i) % mod;
  sort(a + 1, a + n + 1);
  sort(b + 1, b + n + 1);
  sort(c + 1, c + n + 1);
  int cnt = 1;
  for (int i = 2; i <= n; ++i) {
    if (a[i] == a[i - 1]) {
      ++cnt;
    } else {
      ans1 = (ans1 * fact[cnt]) % mod;
      cnt = 1;
    }
  }
  if (cnt > 0) ans1 = (ans1 * fact[cnt]) % mod;
  cnt = 1;
  for (int i = 2; i <= n; ++i) {
    if (b[i] == b[i - 1]) {
      ++cnt;
    } else {
      ans2 = (ans2 * fact[cnt]) % mod;
      cnt = 1;
    }
  }
  if (cnt > 0) ans2 = (ans2 * fact[cnt]) % mod;
  bool can = true;
  for (int i = 2; i <= n; ++i) {
    if (c[i].second < c[i - 1].second) {
      can = false;
      break;
    }
  }
  for (int i = 1; i <= n; ++i) {
    int j = i;
    while (j <= n && c[j].first == c[i].first) ++j;
    --j;
    cnt = 1;
    for (int k = i + 1; k <= j; ++k) {
      if (c[k].second == c[k - 1].second) {
        ++cnt;
      } else {
        ans3 = (ans3 * fact[cnt]) % mod;
        cnt = 1;
      }
    }
    if (cnt > 0) ans3 = (ans3 * fact[cnt]) % mod;
    i = j;
  }
  if (can == false) ans3 = 0;
  long long ans = (ans1 + ans2 - ans3 + mod) % mod;
  ans = (fact[n] - ans + mod) % mod;
  cout << ans << "\n";
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
int n;
long long ans1 = 1, ans2 = 1, ans3 = 1;
long long fact[400400];
int a[400400], b[400400];
pair<int, int> c[400400];
const long long mod = 998244353;
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i] >> b[i];
    c[i] = make_pair(a[i], b[i]);
  }
  fact[0] = 1;
  for (int i = 1; i <= n; ++i) fact[i] = (fact[i - 1] * i) % mod;
  sort(a + 1, a + n + 1);
  sort(b + 1, b + n + 1);
  sort(c + 1, c + n + 1);
  int cnt = 1;
  for (int i = 2; i <= n; ++i) {
    if (a[i] == a[i - 1]) {
      ++cnt;
    } else {
      ans1 = (ans1 * fact[cnt]) % mod;
      cnt = 1;
    }
  }
  if (cnt > 0) ans1 = (ans1 * fact[cnt]) % mod;
  cnt = 1;
  for (int i = 2; i <= n; ++i) {
    if (b[i] == b[i - 1]) {
      ++cnt;
    } else {
      ans2 = (ans2 * fact[cnt]) % mod;
      cnt = 1;
    }
  }
  if (cnt > 0) ans2 = (ans2 * fact[cnt]) % mod;
  bool can = true;
  for (int i = 2; i <= n; ++i) {
    if (c[i].second < c[i - 1].second) {
      can = false;
      break;
    }
  }
  for (int i = 1; i <= n; ++i) {
    int j = i;
    while (j <= n && c[j].first == c[i].first) ++j;
    --j;
    cnt = 1;
    for (int k = i + 1; k <= j; ++k) {
      if (c[k].second == c[k - 1].second) {
        ++cnt;
      } else {
        ans3 = (ans3 * fact[cnt]) % mod;
        cnt = 1;
      }
    }
    if (cnt > 0) ans3 = (ans3 * fact[cnt]) % mod;
    i = j;
  }
  if (can == false) ans3 = 0;
  long long ans = (ans1 + ans2 - ans3 + mod) % mod;
  ans = (fact[n] - ans + mod) % mod;
  cout << ans << "\n";
  return 0;
}

```