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 read(T &k) {
  k = 0;
  int flag = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') flag = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    k = k * 10 + c - '0';
    c = getchar();
  }
  k *= flag;
}
const int N = 300005;
struct node {
  int x, y;
  bool operator<(const node &k) const {
    if (x != k.x) return x < k.x;
    return y < k.y;
  }
} a[N];
int n, m, k, s, t;
long long ans, b[N];
void calc() {
  long long l = 1, xx = 1;
  while (l <= n) {
    long long r = l + 1;
    while (r <= n && a[l].x == a[r].x) ++r;
    xx = xx * b[r - l] % 998244353;
    l = r;
  }
  ans -= xx;
}
void calc1() {
  for (int i = 2; i <= n; ++i)
    if (a[i].y < a[i - 1].y) return;
  long long l = 1, xx = 1;
  while (l <= n) {
    long long r = l + 1;
    while (r <= n && a[l].x == a[r].x && a[l].y == a[r].y) ++r;
    xx = xx * b[r - l] % 998244353;
    l = r;
  }
  ans += xx;
}
int main() {
  read(n);
  for (int i = 1; i <= n; ++i) {
    read(a[i].x);
    read(a[i].y);
  }
  b[0] = 1;
  for (int i = 1; i <= n; ++i) b[i] = b[i - 1] * i % 998244353;
  ans = b[n];
  sort(a + 1, a + 1 + n);
  calc();
  for (int i = 1; i <= n; ++i) swap(a[i].x, a[i].y);
  sort(a + 1, a + 1 + n);
  calc();
  calc1();
  ans = (ans % 998244353 + 998244353) % 998244353;
  printf("%lld\n", ans);
}
