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 mod = 998244353;
const int inf = 1 << 30;
const int maxn = 300000 + 5;
inline int add(int x, int y) {
  x += y;
  return x >= mod ? x -= mod : x;
}
inline int sub(int x, int y) {
  x -= y;
  return x < 0 ? x += mod : x;
}
inline int mul(int x, int y) { return 1ll * x * y % mod; }
inline int qpow(int x, long long n) {
  int r = 1;
  while (n > 0) {
    if (n & 1) r = 1ll * r * x % mod;
    n >>= 1;
    x = 1ll * x * x % mod;
  }
  return r;
}
inline int inv(int x) { return qpow(x, mod - 2); }
int n, f[maxn];
pair<int, int> a[maxn];
int main() {
  f[0] = 1;
  scanf("%d", &n);
  map<int, int> mpa, mpb;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &a[i].first, &a[i].second);
    mpa[a[i].first]++;
    mpb[a[i].second]++;
    f[i] = mul(f[i - 1], i);
  }
  int ans = f[n], tot = 1;
  for (auto& x : mpa) tot = mul(tot, f[x.second]);
  ans = sub(ans, tot);
  tot = 1;
  for (auto& x : mpb) tot = mul(tot, f[x.second]);
  ans = sub(ans, tot);
  sort(a + 1, a + 1 + n);
  int flag = 1;
  for (int i = 2; i <= n; i++) {
    if (a[i].first < a[i - 1].first || a[i].second < a[i - 1].second) {
      flag = 0;
      break;
    }
  }
  if (flag) {
    map<pair<int, int>, int> mp;
    for (int i = 1; i <= n; i++) {
      mp[a[i]]++;
    }
    tot = 1;
    for (auto& x : mp) tot = mul(tot, f[x.second]);
    ans = add(ans, tot);
  }
  printf("%d", ans);
  return 0;
}
