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 N = 3e5 + 10;
const int mod = 998244353;
const int inf = 0x3f3f3f3f;
int fac[N], n;
struct node {
  int a, b;
} g[N];
bool cmp1(const node &o1, const node &o2) { return o1.a < o2.a; }
bool cmp2(const node &o1, const node &o2) { return o1.b < o2.b; }
bool cmp3(const node &o1, const node &o2) {
  if (o1.a == o2.a) return o1.b < o2.b;
  return o1.a < o2.a;
}
int add(int v) { return (v % mod + mod) % mod; }
bool check() {
  for (int i = 2; i <= n; i++) {
    if (g[i].a < g[i - 1].a || g[i].b < g[i - 1].b) return false;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) {
    fac[i] = fac[i - 1] * 1LL * i % mod;
    scanf("%d%d", &g[i].a, &g[i].b);
  }
  sort(g + 1, g + n + 1, cmp1);
  int inv = 1;
  int p = 1;
  while (p <= n) {
    int v = g[p].a, cnt = 0;
    while (p <= n && g[p].a == v) cnt++, p++;
    inv = inv * 1LL * fac[cnt] % mod;
  }
  sort(g + 1, g + n + 1, cmp2);
  p = 1;
  int inv2 = 1;
  while (p <= n) {
    int v = g[p].b, cnt = 0;
    while (p <= n && g[p].b == v) cnt++, p++;
    inv2 = inv2 * 1LL * fac[cnt] % mod;
  }
  int inv3 = 0;
  sort(g + 1, g + n + 1, cmp3);
  if (check()) {
    inv3 = 1;
    p = 1;
    while (p <= n) {
      int v1 = g[p].a, v2 = g[p].b, cnt = 0;
      while (p <= n && g[p].a == v1 && g[p].b == v2) cnt++, p++;
      inv3 = inv3 * 1LL * fac[cnt] % mod;
    }
  }
  inv = (inv + inv2) % mod;
  inv = ((inv - inv3) % mod + mod) % mod;
  inv = ((fac[n] - inv) % mod + mod) % mod;
  printf("%d\n", inv);
  return 0;
}
