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>
#pragma GCC optimize("Ofast")
using namespace std;
const long long inf = 1e18;
const long double pi = 3.141592653589793238;
long long MOD = 998244353;
const long long N = 3e5 + 10;
long long n;
pair<long long, long long> a[N], b[N];
long long fact[N];
bool cmp(pair<long long, long long>& a, pair<long long, long long>& b) {
  return (a.second < b.second) || (a.second == b.second && a.first < b.first);
}
bool cmp2(pair<long long, long long>& a, pair<long long, long long>& b) {
  return (a.first < b.first) || (a.first == b.first && a.second < b.second);
}
void solve() {
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i].first >> a[i].second;
  for (long long i = 0; i < n; i++) b[i] = a[i];
  long long op1 = 1;
  sort(a, a + n, cmp2);
  long long cur = 0;
  for (long long i = 0; i < n; i++) {
    if (i > 0 && a[i].first != a[i - 1].first) {
      op1 *= (fact[cur]);
      op1 %= MOD;
      op1 += MOD;
      op1 %= MOD;
      cur = 1;
    } else
      cur++;
  }
  op1 *= fact[cur];
  op1 %= MOD;
  op1 += MOD;
  op1 %= MOD;
  sort(b, b + n, cmp);
  long long op2 = 1;
  cur = 0;
  for (long long i = 0; i < n; i++) {
    if (i > 0 && b[i].second != b[i - 1].second) {
      op2 *= (fact[cur]);
      op2 %= MOD;
      op2 += MOD;
      op2 %= MOD;
      cur = 1;
    } else
      cur++;
  }
  op2 *= fact[cur];
  op2 %= MOD;
  op2 += MOD;
  op2 %= MOD;
  long long op3 = 1;
  cur = 0;
  for (long long i = 0; i < n; i++) {
    if ((a[i].first != b[i].first) || (a[i].second != b[i].second)) op3 = 0;
    if (i > 0 &&
        ((b[i].second != b[i - 1].second) || (a[i].first != a[i - 1].first))) {
      op3 *= (fact[cur]);
      op3 %= MOD;
      op3 += MOD;
      op3 %= MOD;
      cur = 1;
    } else
      cur++;
  }
  op3 *= fact[cur];
  op3 %= MOD;
  op3 += MOD;
  op3 %= MOD;
  long long ans = fact[n] - (op1 + op2 - op3);
  ans %= MOD;
  ans += MOD;
  ans %= MOD;
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long t = 1;
  fact[0] = 1;
  for (long long i = 1; i < N; i++) {
    fact[i] = fact[i - 1] * i;
    fact[i] %= MOD;
    fact[i] += MOD;
    fact[i] %= MOD;
  }
  while (t--) {
    solve();
  }
}
