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 maxn = 3e5 + 5;
const int Mod = 998244353;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const double e = exp(1);
const double PI = acos(-1);
const double ERR = 1e-10;
struct node {
  int a, b;
} data[maxn];
bool cmp1(node A, node B) { return A.a < B.a; }
bool cmp2(node A, node B) { return A.b < B.b; }
bool cmp3(node A, node B) { return A.a < B.a || (A.a == B.a && A.b < B.b); }
long long fac[maxn];
int main() {
  int n;
  scanf("%d", &n);
  fac[0] = 1;
  for (int i = 1; i <= 300000; i++) {
    fac[i] = fac[i - 1] * 1LL * i % Mod;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &data[i].a, &data[i].b);
  }
  sort(data + 1, data + 1 + n, cmp1);
  long long res1 = 1;
  int cnt = 1;
  for (int i = 2; i <= n + 1; i++) {
    if (data[i].a == data[i - 1].a) {
      cnt++;
    } else {
      res1 = res1 * fac[cnt] % Mod;
      cnt = 1;
    }
  }
  sort(data + 1, data + 1 + n, cmp2);
  long long res2 = 1;
  cnt = 1;
  for (int i = 2; i <= n + 1; i++) {
    if (data[i].b == data[i - 1].b) {
      cnt++;
    } else {
      res2 = res2 * fac[cnt] % Mod;
      cnt = 1;
    }
  }
  sort(data + 1, data + 1 + n, cmp3);
  bool isok = true;
  for (int i = 2; i <= n; i++) {
    if (!(data[i].a >= data[i - 1].a && data[i].b >= data[i - 1].b)) {
      isok = false;
      break;
    }
  }
  long long res3 = (long long)isok;
  cnt = 1;
  for (int i = 2; i <= n + 1; i++) {
    if (data[i].a == data[i - 1].a && data[i].b == data[i - 1].b) {
      cnt++;
    } else {
      res3 = res3 * fac[cnt] % Mod;
      cnt = 1;
    }
  }
  long long ans = fac[n];
  ans = (ans - res1 + Mod) % Mod;
  ans = (ans - res2 + Mod) % Mod;
  ans = (ans + res3) % Mod;
  printf("%lld\n", ans);
  return 0;
}
