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 T sqr(T x) {
  return x * x;
}
const double EPS = 1e-6;
const int INF = 0x3fffffff;
const long long LINF = INF * 1ll * INF;
const double PI = acos(-1.0);
pair<int, int> p[300005];
int cnt1[300005], cnt2[300005];
map<pair<int, int>, int> cnt3;
long long fac[300005];
bool cmp(pair<int, int> x, pair<int, int> y) {
  if (x.first == y.first) return x.second < y.second;
  return x.first < y.first;
}
int main() {
  int n;
  scanf("%d", &n);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) {
    fac[i] = fac[i - 1] * i % 998244353;
    scanf("%d%d", &p[i].first, &p[i].second);
    cnt1[p[i].first]++;
    cnt2[p[i].second]++;
    cnt3[p[i]]++;
  }
  sort(p + 1, p + 1 + n, cmp);
  long long res1 = 1, res2 = 1;
  for (int i = 0; i <= n; i++) {
    if (cnt1[i] > 1) res1 = res1 * fac[cnt1[i]] % 998244353;
    if (cnt2[i] > 1) res2 = res2 * fac[cnt2[i]] % 998244353;
  }
  bool flag = true;
  for (int i = 1; i < n; i++) {
    if (p[i].second > p[i + 1].second) {
      flag = false;
      break;
    }
  }
  long long ans;
  if (flag) {
    long long res3 = 1;
    for (auto &it : cnt3) {
      if (it.second > 1) res3 = res3 * fac[it.second] % 998244353;
    }
    ans = ((fac[n] - res1 - res2 + res3) % 998244353 + 998244353) % 998244353;
  } else
    ans = ((fac[n] - res1 - res2) % 998244353 + 998244353) % 998244353;
  printf("%lld\n", ans);
  return 0;
}
