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;
template <int MOD>
struct ModInt {
  static const int Mod = MOD;
  unsigned x;
  ModInt() : x(0) {}
  ModInt(signed sig) {
    int sigt = sig % MOD;
    if (sigt < 0) sigt += MOD;
    x = sigt;
  }
  ModInt(signed long long sig) {
    int sigt = sig % MOD;
    if (sigt < 0) sigt += MOD;
    x = sigt;
  }
  int get() const { return (int)x; }
  ModInt &operator+=(ModInt that) {
    if ((x += that.x) >= MOD) x -= MOD;
    return *this;
  }
  ModInt &operator-=(ModInt that) {
    if ((x += MOD - that.x) >= MOD) x -= MOD;
    return *this;
  }
  ModInt &operator*=(ModInt that) {
    x = (unsigned long long)x * that.x % MOD;
    return *this;
  }
  ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }
  ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
  ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
  ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
  ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }
  ModInt inverse() const {
    signed a = x, b = MOD, u = 1, v = 0;
    while (b) {
      signed t = a / b;
      a -= t * b;
      std::swap(a, b);
      u -= t * v;
      std::swap(u, v);
    }
    if (u < 0) u += Mod;
    ModInt res;
    res.x = (unsigned)u;
    return res;
  }
  bool operator==(ModInt that) const { return x == that.x; }
  bool operator!=(ModInt that) const { return x != that.x; }
  ModInt operator-() const {
    ModInt t;
    t.x = x == 0 ? 0 : Mod - x;
    return t;
  }
};
template <int MOD>
ModInt<MOD> operator^(ModInt<MOD> a, unsigned long long k) {
  ModInt<MOD> r = 1;
  while (k) {
    if (k & 1) r *= a;
    a *= a;
    k >>= 1;
  }
  return r;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<ModInt<998244353>> fact(n + 1, ModInt<998244353>(1));
  for (int i = 1; i <= n; ++i) {
    fact[i] = ModInt<998244353>(i) * fact[i - 1];
  }
  vector<pair<int, int>> s(n);
  vector<int> fa(n + 1, 0), fb(n + 1, 0);
  for (int i = 0; i < n; ++i) {
    cin >> s[i].first >> s[i].second;
    fa[s[i].first]++;
    fb[s[i].second]++;
  }
  ModInt<998244353> spa = ModInt<998244353>(1), spb = ModInt<998244353>(1);
  for (int i = 1; i <= n; ++i) {
    spa *= fact[fa[i]];
    spb *= fact[fb[i]];
  }
  sort(s.begin(), s.end());
  ModInt<998244353> spab = ModInt<998244353>(1);
  for (int i = 0; i < n - 1; ++i) {
    if (s[i].second > s[i + 1].second) spab = ModInt<998244353>(0);
  }
  for (int i = 0, j = 0; i < n; i = j) {
    j = i + 1;
    while (j < n && s[j] == s[i]) j++;
    spab *= fact[j - i];
  }
  ModInt<998244353> ans = fact[n] - spa - spb + spab;
  cout << ans.get();
}
