#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 998244353;
int n;
vector<pair<int, int> > v, v2, vtroll;
long long int fat[300300];
map<pair<int, int>, int> freq, freq2;
long long int fastxp(long long int base, long long int exp) {
  long long int ans = 1;
  while (exp) {
    if (exp % 1) ans *= base;
    base *= base;
    base %= MOD;
    ans %= MOD;
    exp /= 2;
  }
  return ans;
}
long long int invmod(long long int base) { return fastxp(base, MOD - 1); }
bool check(vector<pair<int, int> > cara) {
  bool flag1 = true;
  bool flag2 = true;
  for (int i = 1; i < n; i++) {
    if (cara[i].first < cara[i - 1].first) flag1 = false;
    if (cara[i].second < cara[i - 1].second) flag2 = false;
  }
  return flag1 | flag2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  fat[0] = 1ll;
  fat[1] = 1ll;
  for (long long int i = 2; i < 300300; i++) fat[i] = (i * fat[i - 1]) % MOD;
  for (int i = 1; i <= n; i++) {
    int a, b;
    cin >> a >> b;
    v.push_back({a, b});
    v2.push_back({b, a});
    vtroll.push_back({a, b});
    freq[{a, b}]++;
    freq2[{b, a}]++;
  }
  sort(v.begin(), v.end());
  int ini = 0;
  long long int badfirst = 1;
  while (ini < n) {
    long long int val = v[ini].first;
    long long int cnt = 1;
    set<pair<int, int> > lol;
    lol.insert(v[ini]);
    while (ini + 1 < n && v[ini + 1].first == v[ini].first) {
      ini++;
      cnt++;
      lol.insert(v[ini]);
    }
    badfirst *= fat[cnt];
    badfirst %= MOD;
    for (auto x : lol) badfirst *= invmod(fat[freq[x]]);
    ini++;
  }
  ini = 0;
  long long int badsecond = 1;
  sort(v2.begin(), v2.end());
  while (ini < n) {
    long long int val = v2[ini].first;
    long long int cnt = 1;
    set<pair<int, int> > lol;
    lol.insert(v[ini]);
    while (ini + 1 < n && v2[ini + 1].first == v2[ini].first) {
      ini++;
      cnt++;
      lol.insert(v[ini]);
    }
    badsecond *= fat[cnt];
    badsecond %= MOD;
    for (auto x : lol) badsecond *= invmod(fat[freq2[x]]);
    ini++;
  }
  ini = 0;
  long long int badboth = 1;
  for (int i = 1; i < n; i++) {
    if (v[i].second < v[i - 1].second) badboth = 0;
  }
  while (ini < n) {
    pair<int, int> val = v[ini];
    long long int cnt = 1;
    set<pair<int, int> > lol;
    lol.insert(v[ini]);
    while (ini + 1 < n && v[ini + 1] == v[ini]) {
      ini++;
      cnt++;
      lol.insert(v[ini]);
    }
    badboth *= fat[cnt];
    badboth %= MOD;
    for (auto x : lol) badboth *= invmod(fat[freq[x]]);
    ini++;
  }
  long long int ans = fat[n];
  long long int bad = badfirst + badsecond - badboth;
  bad %= MOD;
  if (bad < 0) bad += MOD;
  ans -= bad;
  if (ans < 0) ans += MOD;
  cout << ans << endl;
}

### 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;
const int INF = 0x3f3f3f3f;
const int MOD = 998244353;
int n;
vector<pair<int, int> > v, v2, vtroll;
long long int fat[300300];
map<pair<int, int>, int> freq, freq2;
long long int fastxp(long long int base, long long int exp) {
  long long int ans = 1;
  while (exp) {
    if (exp % 1) ans *= base;
    base *= base;
    base %= MOD;
    ans %= MOD;
    exp /= 2;
  }
  return ans;
}
long long int invmod(long long int base) { return fastxp(base, MOD - 1); }
bool check(vector<pair<int, int> > cara) {
  bool flag1 = true;
  bool flag2 = true;
  for (int i = 1; i < n; i++) {
    if (cara[i].first < cara[i - 1].first) flag1 = false;
    if (cara[i].second < cara[i - 1].second) flag2 = false;
  }
  return flag1 | flag2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  fat[0] = 1ll;
  fat[1] = 1ll;
  for (long long int i = 2; i < 300300; i++) fat[i] = (i * fat[i - 1]) % MOD;
  for (int i = 1; i <= n; i++) {
    int a, b;
    cin >> a >> b;
    v.push_back({a, b});
    v2.push_back({b, a});
    vtroll.push_back({a, b});
    freq[{a, b}]++;
    freq2[{b, a}]++;
  }
  sort(v.begin(), v.end());
  int ini = 0;
  long long int badfirst = 1;
  while (ini < n) {
    long long int val = v[ini].first;
    long long int cnt = 1;
    set<pair<int, int> > lol;
    lol.insert(v[ini]);
    while (ini + 1 < n && v[ini + 1].first == v[ini].first) {
      ini++;
      cnt++;
      lol.insert(v[ini]);
    }
    badfirst *= fat[cnt];
    badfirst %= MOD;
    for (auto x : lol) badfirst *= invmod(fat[freq[x]]);
    ini++;
  }
  ini = 0;
  long long int badsecond = 1;
  sort(v2.begin(), v2.end());
  while (ini < n) {
    long long int val = v2[ini].first;
    long long int cnt = 1;
    set<pair<int, int> > lol;
    lol.insert(v[ini]);
    while (ini + 1 < n && v2[ini + 1].first == v2[ini].first) {
      ini++;
      cnt++;
      lol.insert(v[ini]);
    }
    badsecond *= fat[cnt];
    badsecond %= MOD;
    for (auto x : lol) badsecond *= invmod(fat[freq2[x]]);
    ini++;
  }
  ini = 0;
  long long int badboth = 1;
  for (int i = 1; i < n; i++) {
    if (v[i].second < v[i - 1].second) badboth = 0;
  }
  while (ini < n) {
    pair<int, int> val = v[ini];
    long long int cnt = 1;
    set<pair<int, int> > lol;
    lol.insert(v[ini]);
    while (ini + 1 < n && v[ini + 1] == v[ini]) {
      ini++;
      cnt++;
      lol.insert(v[ini]);
    }
    badboth *= fat[cnt];
    badboth %= MOD;
    for (auto x : lol) badboth *= invmod(fat[freq[x]]);
    ini++;
  }
  long long int ans = fat[n];
  long long int bad = badfirst + badsecond - badboth;
  bad %= MOD;
  if (bad < 0) bad += MOD;
  ans -= bad;
  if (ans < 0) ans += MOD;
  cout << ans << endl;
}

```