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. 
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
const int maxn = 300005;
struct Item {
  int a, b;
} item[maxn];
long long fac[maxn];
void GCD(long long a, long long b, long long &d, long long &x, long long &y) {
  if (!b) {
    d = a, x = 1, y = 0;
  } else {
    GCD(b, a % b, d, y, x);
    y -= x * (a / b);
  }
}
long long inv(long long a, long long n) {
  long long d, x, y;
  GCD(a, n, d, x, y);
  return d == 1 ? (x + n) % n : -1;
}
void get_fact() {
  fac[0] = 1;
  for (long long i = 1; i < maxn; i++) fac[i] = fac[i - 1] * i % MOD;
}
bool cmpa(const Item &a, const Item &b) {
  return (a.a == b.a ? a.b < b.b : a.a < b.a);
}
bool cmpb(const Item &a, const Item &b) {
  return (a.b == b.b ? a.a < b.a : a.b < b.b);
}
long long solve(int n) {
  long long ans = 1;
  for (int i = 1; i <= n; i++) {
    int len = 0;
    for (int j = i; j <= n; j++) {
      if (i == j || item[j].a == item[j - 1].a) {
        len++;
      } else {
        break;
      }
    }
    ans = ans * fac[len] % MOD;
    i += len - 1;
  }
  return ans;
}
long long solve1(int n) {
  long long ans = 1;
  for (int i = 1; i <= n; i++) {
    int len = 0;
    for (int j = i; j <= n; j++) {
      if (i == j || item[j].b == item[j - 1].b) {
        len++;
      } else {
        break;
      }
    }
    ans = ans * fac[len] % MOD;
    i += len - 1;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  get_fact();
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> item[i].a >> item[i].b;
  }
  long long ans = 0;
  sort(item + 1, item + n + 1, cmpa);
  ans = solve(n);
  sort(item + 1, item + n + 1, cmpb);
  ans = (ans + solve1(n)) % MOD;
  long long ok = 1;
  for (int i = 2; i <= n; i++) {
    if (item[i].a < item[i - 1].a) {
      ok = 0;
      break;
    }
  }
  long long sum = 1;
  if (ok) {
    for (int i = 1; i <= n; i++) {
      int len = 0;
      for (int j = i; j <= n; j++) {
        if (i == j ||
            item[j].b == item[j - 1].b && item[j].a == item[j - 1].a) {
          len++;
        } else {
          break;
        }
      }
      sum = sum * fac[len] % MOD;
      i += len - 1;
    }
    ans = ((ans - sum) % MOD + MOD) % MOD;
  }
  cout << ((fac[n] - ans) % MOD + MOD) % MOD << endl;
  return 0;
}
