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 <class T>
inline void rd(T &x) {
  char ch;
  x = 0;
  bool fl = false;
  while (!isdigit(ch = getchar())) (ch == '-') && (fl = true);
  for (x = (ch ^ '0'); isdigit(ch = getchar()); x = x * 10 + (ch ^ '0'))
    ;
  (fl == true) && (x = -x);
}
template <class T>
inline void output(T x) {
  if (x / 10) output(x / 10);
  putchar(x % 10 + '0');
}
template <class T>
inline void ot(T x) {
  if (x < 0) putchar('-'), x = -x;
  output(x);
  putchar(' ');
}
template <class T>
inline void prt(T a[], int st, int nd) {
  for (int i = st; i <= nd; ++i) ot(a[i]);
  putchar('\n');
}
namespace Modulo {
const int mod = 998244353;
inline int ad(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
inline int sub(int x, int y) { return ad(x, mod - y); }
inline int mul(int x, int y) { return (long long)x * y % mod; }
inline void inc(int &x, int y) { x = ad(x, y); }
inline void inc2(int &x, int y) { x = mul(x, y); }
inline int qm(int x, int y = mod - 2) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = mul(x, ret);
    x = mul(x, x);
    y >>= 1;
  }
  return ret;
}
template <class... Args>
inline int ad(const int a, const int b, const Args &...args) {
  return ad(ad(a, b), args...);
}
template <class... Args>
inline int mul(const int a, const int b, const Args &...args) {
  return mul(mul(a, b), args...);
}
}  // namespace Modulo
using namespace std;
pair<int, int> p[300007];
const long long mod = 998244353LL;
long long fac[300007];
class Task {
 public:
  void solve(istream &in, ostream &out) {
    fac[0] = 1;
    for (long long i = 1; i != 300005; i++) fac[i] = fac[i - 1] * i % mod;
    int n;
    in >> n;
    for (int i = 0; i != n; i++) in >> p[i].first >> p[i].second;
    sort(p, p + n);
    long long ans = fac[n], now = 0, ts = 1, num = 0;
    for (int i = 0; i != n; i++) {
      if (p[i].first != now)
        (ts *= fac[num]) %= mod, num = 1, now = p[i].first;
      else
        num++;
    }
    (ts *= fac[num]) %= mod, ans -= ts;
    bool c = true;
    for (int i = 0; i != n - 1 && c; i++) {
      if (p[i].first > p[i + 1].first || p[i].second > p[i + 1].second) {
        c = false;
      }
    }
    if (c) {
      ts = 1, num = 0;
      for (int i = 0; i != n; i++) {
        if (i == 0 || p[i] != p[i - 1])
          (ts *= fac[num]) %= mod, num = 1;
        else
          num++;
      }
      (ts *= fac[num]) %= mod, ans += ts, num = 1;
    }
    sort(p, p + n, [&](pair<int, int> &p1, pair<int, int> &p2) {
      return p1.second < p2.second;
    });
    now = 0, ts = 1, num = 0;
    for (int i = 0; i != n; i++) {
      if (p[i].second != now)
        (ts *= fac[num]) %= mod, num = 1, now = p[i].second;
      else
        num++;
    }
    (ts *= fac[num]) %= mod, ans -= ts;
    out << (mod + ans % mod) % mod;
  }
};
int main() {
  ios::sync_with_stdio(false);
  Task solver;
  std::istream &in(std::cin);
  std::ostream &out(std::cout);
  solver.solve(in, out);
  return 0;
}
