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 long double pi = acos((long double)-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const int maxn = 1e6 + 5;
const int maxm = 1e6 + 5;
const long long mod = 998244353;
void in(initializer_list<int *> li) {
  for (auto ptr = li.begin(); ptr != li.end(); ptr++) scanf("%d", *ptr);
}
void in(initializer_list<long long *> li) {
  for (auto ptr = li.begin(); ptr != li.end(); ptr++) scanf("%lld", *ptr);
}
void in(initializer_list<double *> li) {
  for (auto ptr = li.begin(); ptr != li.end(); ptr++) scanf("%lf", *ptr);
}
void out(initializer_list<int> li) {
  for (auto ptr = li.begin(); ptr != li.end(); ptr++) printf("%d ", *ptr);
}
void outln(initializer_list<int> li) {
  for (auto ptr = li.begin(); ptr != li.end(); ptr++) printf("%d\n", *ptr);
}
void out(initializer_list<long long> li) {
  for (auto ptr = li.begin(); ptr != li.end(); ptr++) printf("%lld ", *ptr);
}
void outln(initializer_list<long long> li) {
  for (auto ptr = li.begin(); ptr != li.end(); ptr++) printf("%lld\n", *ptr);
}
int in(int &a, int &b, int &c, int &d) {
  return scanf("%d%d%d%d", &a, &b, &c, &d);
}
int in(int &a, int &b, int &c) { return scanf("%d%d%d", &a, &b, &c); }
int in(int &a, int &b) { return scanf("%d%d", &a, &b); }
int in(long long &a) { return scanf("%lld", &a); }
int in(int &a) { return scanf("%d", &a); }
int in(char *s) { return scanf("%s", s); }
int in(char &c) { return scanf("%c", &c); }
void out(int a, bool ln) { printf("%d%c", a, " \n"[ln]); }
void out(long long a, bool ln) { printf("%lld%c", a, " \n"[ln]); }
void out(double a, int digit, bool ln) {
  printf("%.*f%c", digit, a, " \n"[ln]);
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
map<int, long long> ha, hb;
map<pair<int, int>, long long> hab;
map<int, long long> temp;
long long ans = 0, suma = 1, sumb = 1, sumab = 1;
struct node {
  int a, b;
} no[maxn];
bool cmp1(node a, node b) {
  if (a.a == b.a) return a.b < b.b;
  return a.a < b.a;
}
long long qpl[maxn];
void init(int n) {
  qpl[1] = 1;
  long long ans = 1;
  for (int i = 2; i <= n; ++i) {
    ans *= i;
    ans %= mod;
    qpl[i] = ans;
  }
}
int main() {
  int n;
  in(n);
  init(n);
  for (int i = 0; i < n; ++i) {
    in(no[i].a, no[i].b);
    ha[no[i].a]++, hb[no[i].b]++;
    hab[make_pair(no[i].a, no[i].b)]++;
  }
  map<int, long long>::iterator it = ha.begin();
  for (; it != ha.end(); it++) {
    suma *= qpl[it->second];
    suma %= mod;
  }
  it = hb.begin();
  for (; it != hb.end(); it++) {
    sumb *= qpl[it->second];
    sumb %= mod;
  }
  ans = (qpl[n]) % mod;
  sort(no, no + n, cmp1);
  bool flag = 1;
  for (int i = 1; i < n; ++i) {
    if (no[i].b < no[i - 1].b) {
      flag = 0;
      break;
    }
  }
  if (!flag) {
    ans = (ans - suma - sumb + mod * 3) % mod;
    cout << ans << "\n";
  } else {
    map<pair<int, int>, long long>::iterator it = hab.begin();
    for (; it != hab.end(); it++) {
      sumab *= qpl[it->second];
      sumab %= mod;
    }
    ans = (ans + sumab) % mod;
    ans = (ans - suma - sumb + mod * 3) % mod;
    cout << ans << "\n";
  }
  return 0;
}
