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;
using ll = long long int;
using int64 = long long int;
template <typename T>
void chmax(T &a, T b) {
  a = max(a, b);
}
template <typename T>
void chmin(T &a, T b) {
  a = min(a, b);
}
template <typename T>
void chadd(T &a, T b) {
  a = a + b;
}
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
const ll INF = 1001001001001001LL;
const ll MOD = 998244353LL;
int main() {
  int N;
  scanf("%d", &N);
  vector<ll> perm(N + 1);
  perm[0] = 1;
  for (int i = 1; i <= N; i++) perm[i] = (perm[i - 1] * i) % MOD;
  map<pair<int, int>, int> cnt;
  for (int i = 0; i < N; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    cnt[make_pair(a, b)]++;
  }
  vector<pair<int, int> > P;
  for (auto e : cnt) P.emplace_back(e.first);
  bool both_sorted = true;
  for (size_t i = 1; i < P.size(); i++) {
    both_sorted &= (P[i - 1].first <= P[i].first);
    both_sorted &= (P[i - 1].second <= P[i].second);
  }
  ll ans = 0;
  {
    ll sum = 1;
    map<int, int> cntA;
    for (auto e : P) {
      int val = e.first;
      int num = cnt[e];
      cntA[val] += num;
    }
    for (auto e : cntA) {
      (sum *= perm[e.second]) %= MOD;
    }
    fprintf(stderr, "sum = %lld\n", sum);
    ans += sum;
    ans %= MOD;
  }
  {
    ll sum = 1;
    map<int, int> cntB;
    for (auto e : P) {
      int val = e.second;
      int num = cnt[e];
      cntB[val] += num;
    }
    for (auto e : cntB) {
      (sum *= perm[e.second]) %= MOD;
    }
    fprintf(stderr, "sum = %lld\n", sum);
    ans += sum;
    ans %= MOD;
  }
  if (both_sorted) {
    ll sum = 1;
    for (auto e : cnt) {
      (sum *= perm[e.second]) %= MOD;
    }
    fprintf(stderr, "sum = %lld\n", sum);
    ans += MOD - sum;
    ans %= MOD;
  }
  ans = perm[N] + MOD - ans;
  ans %= MOD;
  cout << ans << endl;
  return 0;
}
