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 = 1e9 + 7;
const int infi = 1147483600;
const long long infl = 1e18 + 5;
const char ENDL = '\n';
const long long MOD = 998244353;
struct COMB {
  int L;
  vector<long long> kaijo, gyaku;
  long long mod;
  void __init__(int sz, long long md) {
    L = sz;
    mod = md;
    kaijo.resize(L + 1, 0);
    gyaku.resize(L + 1, 0);
    kaijo[0] = 1;
    for (long long i = (long long)1; i < (long long)L + 1; i++)
      kaijo[i] = (kaijo[i - 1] * i) % mod;
    gyaku[L] = powmod(kaijo[L], mod - 2, mod);
    for (long long i = 0; i < (long long)L; i++)
      gyaku[L - 1 - i] = gyaku[L - i] * (L - i) % mod;
  }
  long long powmod(long long x, long long y, long long mod) {
    long long ans = 1;
    while (y > 0) {
      if (y & 1) {
        ans *= x;
        ans %= mod;
      }
      x = x * x % mod;
      y >>= 1;
    }
    return ans;
  }
  long long calc(int n, int k) {
    if (n > L) return -1;
    return (((kaijo[n] * gyaku[k]) % mod) * gyaku[n - k]) % mod;
  }
  long long kai(int i) { return kaijo[i]; }
  long long gya(int i) { return gyaku[i]; }
};
bool comp1(pair<long long, long long> a, pair<long long, long long> b) {
  return a.first < b.first;
}
bool comp2(pair<long long, long long> a, pair<long long, long long> b) {
  return a.second < b.second;
}
bool comp3(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first)
    return a.second < b.second;
  else
    return a.first < b.first;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n;
  cin >> n;
  vector<pair<long long, long long> > s(n);
  for (long long i = 0; i < (long long)n; i++) {
    long long a, b;
    cin >> a >> b;
    s[i] = make_pair(a, b);
  }
  COMB comb;
  comb.__init__(n, MOD);
  long long sa = 1;
  sort((s).begin(), (s).end(), comp1);
  long long bef = s[0].first;
  long long cnt = 1;
  for (long long i = (long long)1; i < (long long)n; i++) {
    if (s[i].first == bef) {
      cnt++;
    } else {
      sa *= comb.kai(cnt);
      sa %= MOD;
      cnt = 1;
      bef = s[i].first;
    }
  }
  sa *= comb.kai(cnt);
  sa %= MOD;
  long long sb = 1;
  sort((s).begin(), (s).end(), comp2);
  bef = s[0].second;
  cnt = 1;
  for (long long i = (long long)1; i < (long long)n; i++) {
    if (s[i].second == bef) {
      cnt++;
    } else {
      sb *= comb.kai(cnt);
      sb %= MOD;
      cnt = 1;
      bef = s[i].second;
    }
  }
  sb *= comb.kai(cnt);
  sb %= MOD;
  sort((s).begin(), (s).end(), comp3);
  long long ss = 1;
  long long befa = s[0].first, befb = s[0].second;
  cnt = 1;
  for (long long i = (long long)1; i < (long long)n; i++) {
    if (s[i - 1].first > s[i].first || s[i - 1].second > s[i].second) {
      ss = 0;
      break;
    }
    if (s[i].first == befa && s[i].second == befb) {
      cnt++;
    } else {
      ss *= comb.kai(cnt);
      ss %= MOD;
      cnt = 1;
      befa = s[i].first;
      befb = s[i].second;
    }
  }
  ss *= comb.kai(cnt);
  ss %= MOD;
  cout << (2 * MOD + comb.kai(n) - sa - sb + ss) % MOD << ENDL;
}
