#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18;
const double PI = acos(-1.0);
const long long MAX_N = 300005;
const long long MOD = 998244353;
long long n;
vector<pair<long long, long long>> va, vb;
long long fact[MAX_N];
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  long long a, b;
  for (long long i = 0; i < n; i++) {
    cin >> a >> b;
    va.push_back({a, b});
    vb.push_back({b, a});
  }
  sort(va.begin(), va.end());
  sort(vb.begin(), vb.end());
  fact[0] = 1;
  for (long long i = 1; i <= n; i++) {
    fact[i] = (fact[i - 1] * i) % MOD;
  }
  long long cnt1 = 1, cnt2 = 1, cnt3 = 1;
  long long cnt = 0;
  for (long long i = 0; i < n; i++) {
    if (i == 0 || va[i].first == va[i - 1].first) {
      cnt++;
    } else {
      cnt1 = (cnt1 * fact[cnt]) % MOD;
      cnt = 1;
    }
  }
  cnt1 = (cnt1 * fact[cnt]) % MOD;
  cnt = 0;
  for (long long i = 0; i < n; i++) {
    if (i == 0 || vb[i].first == vb[i - 1].first) {
      cnt++;
    } else {
      cnt2 = (cnt2 * fact[cnt]) % MOD;
      cnt = 1;
    }
  }
  cnt2 = (cnt2 * fact[cnt]) % MOD;
  cnt = 0;
  for (long long i = 0; i < n; i++) {
    if (i == 0 || va[i] == va[i - 1]) {
      cnt++;
    } else {
      cnt3 = (cnt3 * fact[cnt]) % MOD;
      cnt = 1;
    }
    if (i > 0 && va[i].second < va[i - 1].second) {
      cnt = 0;
      cnt3 = 0;
      break;
    }
  }
  cnt3 = (cnt3 * fact[cnt]) % MOD;
  long long res = fact[n];
  res -= cnt1;
  res += MOD;
  res %= MOD;
  res -= cnt2;
  res += MOD;
  res %= MOD;
  res += cnt3;
  res %= MOD;
  cout << res;
  return 0;
}

### Prompt

Your task is to create a cpp solution 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;
const long long INF = 1e18;
const double PI = acos(-1.0);
const long long MAX_N = 300005;
const long long MOD = 998244353;
long long n;
vector<pair<long long, long long>> va, vb;
long long fact[MAX_N];
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  long long a, b;
  for (long long i = 0; i < n; i++) {
    cin >> a >> b;
    va.push_back({a, b});
    vb.push_back({b, a});
  }
  sort(va.begin(), va.end());
  sort(vb.begin(), vb.end());
  fact[0] = 1;
  for (long long i = 1; i <= n; i++) {
    fact[i] = (fact[i - 1] * i) % MOD;
  }
  long long cnt1 = 1, cnt2 = 1, cnt3 = 1;
  long long cnt = 0;
  for (long long i = 0; i < n; i++) {
    if (i == 0 || va[i].first == va[i - 1].first) {
      cnt++;
    } else {
      cnt1 = (cnt1 * fact[cnt]) % MOD;
      cnt = 1;
    }
  }
  cnt1 = (cnt1 * fact[cnt]) % MOD;
  cnt = 0;
  for (long long i = 0; i < n; i++) {
    if (i == 0 || vb[i].first == vb[i - 1].first) {
      cnt++;
    } else {
      cnt2 = (cnt2 * fact[cnt]) % MOD;
      cnt = 1;
    }
  }
  cnt2 = (cnt2 * fact[cnt]) % MOD;
  cnt = 0;
  for (long long i = 0; i < n; i++) {
    if (i == 0 || va[i] == va[i - 1]) {
      cnt++;
    } else {
      cnt3 = (cnt3 * fact[cnt]) % MOD;
      cnt = 1;
    }
    if (i > 0 && va[i].second < va[i - 1].second) {
      cnt = 0;
      cnt3 = 0;
      break;
    }
  }
  cnt3 = (cnt3 * fact[cnt]) % MOD;
  long long res = fact[n];
  res -= cnt1;
  res += MOD;
  res %= MOD;
  res -= cnt2;
  res += MOD;
  res %= MOD;
  res += cnt3;
  res %= MOD;
  cout << res;
  return 0;
}

```