#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 1;
const long long LINF = 1e17;
const int MOD = 998244353;
const int MAKS = 400100;
int n;
bool boo;
long long fak[MAKS], ans[3], cnt[2][MAKS], cntt;
vector<long long> lst[MAKS];
pair<int, int> p[MAKS];
int main() {
  scanf("%d", &n);
  fak[0] = 1;
  for (int i = 1; i <= n; i++) {
    fak[i] = fak[i - 1] * i;
    fak[i] %= MOD;
  }
  for (int i = 0; i < 3; i++) ans[i] = 1;
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &p[i].first, &p[i].second);
    cnt[0][p[i].first]++;
    cnt[1][p[i].second]++;
    lst[p[i].first].push_back(p[i].second);
  }
  for (int j = 0; j < 2; j++) {
    for (int i = 1; i <= n; i++) {
      ans[j] *= fak[cnt[j][i]];
      ans[j] %= MOD;
    }
  }
  long long last = 0;
  for (int i = 1; i <= n; i++) {
    if (lst[i].size() == 0) continue;
    sort(lst[i].begin(), lst[i].end());
    if (boo && lst[i].front() < last) {
      ans[2] = 0;
      break;
    }
    boo = true;
    cntt = 0;
    for (int j = 0; j < lst[i].size(); j++) {
      if (j == 0 || lst[i][j] == lst[i][j - 1])
        cntt++;
      else {
        ans[2] *= fak[cntt];
        ans[2] %= MOD;
        cntt = 1;
      }
    }
    ans[2] *= fak[cntt];
    ans[2] %= MOD;
    last = lst[i].back();
  }
  long long temp = ans[0] + ans[1];
  temp -= ans[2];
  temp = fak[n] - temp;
  while (temp < 0) temp += MOD;
  temp %= MOD;
  printf("%lld\n", temp);
}

### 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 = 1e9 + 1;
const long long LINF = 1e17;
const int MOD = 998244353;
const int MAKS = 400100;
int n;
bool boo;
long long fak[MAKS], ans[3], cnt[2][MAKS], cntt;
vector<long long> lst[MAKS];
pair<int, int> p[MAKS];
int main() {
  scanf("%d", &n);
  fak[0] = 1;
  for (int i = 1; i <= n; i++) {
    fak[i] = fak[i - 1] * i;
    fak[i] %= MOD;
  }
  for (int i = 0; i < 3; i++) ans[i] = 1;
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &p[i].first, &p[i].second);
    cnt[0][p[i].first]++;
    cnt[1][p[i].second]++;
    lst[p[i].first].push_back(p[i].second);
  }
  for (int j = 0; j < 2; j++) {
    for (int i = 1; i <= n; i++) {
      ans[j] *= fak[cnt[j][i]];
      ans[j] %= MOD;
    }
  }
  long long last = 0;
  for (int i = 1; i <= n; i++) {
    if (lst[i].size() == 0) continue;
    sort(lst[i].begin(), lst[i].end());
    if (boo && lst[i].front() < last) {
      ans[2] = 0;
      break;
    }
    boo = true;
    cntt = 0;
    for (int j = 0; j < lst[i].size(); j++) {
      if (j == 0 || lst[i][j] == lst[i][j - 1])
        cntt++;
      else {
        ans[2] *= fak[cntt];
        ans[2] %= MOD;
        cntt = 1;
      }
    }
    ans[2] *= fak[cntt];
    ans[2] %= MOD;
    last = lst[i].back();
  }
  long long temp = ans[0] + ans[1];
  temp -= ans[2];
  temp = fak[n] - temp;
  while (temp < 0) temp += MOD;
  temp %= MOD;
  printf("%lld\n", temp);
}

```