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 int maxn = 3e5 + 10, mod = 998244353;
long long p[maxn];
void add(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
  if (x < 0) x += mod;
}
int X[maxn], Y[maxn];
int mx[maxn], mn[maxn], vis[maxn];
vector<int> G[maxn];
set<int> s;
int main() {
  p[0] = 1;
  for (int i = 1; i < maxn; i++) p[i] = p[i - 1] * i % mod;
  int n, x, y;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) mn[i] = n + 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &x, &y);
    X[x]++;
    Y[y]++;
    mx[x] = max(mx[x], y);
    mn[x] = min(mn[x], y);
    G[x].push_back(y);
  }
  long long ans = p[n], t1 = 1, t2 = 1;
  for (int i = 1; i <= n; i++) {
    t1 = t1 * p[X[i]] % mod;
    t2 = t2 * p[Y[i]] % mod;
  }
  add(ans, -t1);
  add(ans, -t2);
  int ok = 1;
  for (int i = 1; i <= n && ok; i++) {
    if (mn[i] < mx[i - 1])
      ok = 0;
    else
      mx[i] = max(mx[i], mx[i - 1]);
  }
  if (ok) {
    long long tmp = 1;
    for (int i = 1; i <= n; i++) {
      if (G[i].empty()) continue;
      s.clear();
      for (auto v : G[i]) vis[v]++, s.insert(v);
      for (auto it : s) tmp = tmp * p[vis[it]] % mod;
      for (auto it : s) vis[it] = 0;
    }
    add(ans, tmp);
  }
  cout << ans;
}
