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 = 5e5 + 100;
const long long inf = 1e18 + 10;
long long mod = 998244353;
long long sui[maxn];
int n, m, num[maxn];
struct node {
  int x, y, id;
} a[maxn], b[maxn];
bool cmp1(node x, node y) { return x.x < y.x; };
bool cmp2(node x, node y) { return x.y < y.y; };
bool cmp3(node x, node y) {
  if (x.x == y.x)
    return x.y < y.y;
  else
    return x.x < y.x;
}
long long judge() {
  long long ans = 1, cnt = 1;
  sort(a + 1, a + 1 + n, cmp3);
  for (int i = 1; i < n; i++)
    if (a[i].y > a[i + 1].y) return 0;
  for (int i = 1; i <= n; i++) {
    if (a[i].x == a[i - 1].x && a[i].y == a[i - 1].y)
      cnt++;
    else {
      ans = ans * sui[cnt] % mod;
      cnt = 1;
    }
  }
  ans = (ans * sui[cnt]) % mod;
  return ans;
}
long long slove1() {
  int num = 0, pre = a[1].x;
  long long ans = 1;
  for (int i = 1; i <= n; i++) {
    if (a[i].x == pre)
      num++;
    else {
      pre = a[i].x;
      ans = (ans * sui[num]) % mod;
      num = 1;
    }
  }
  ans = (ans * sui[num]) % mod;
  return ans;
}
long long slove2() {
  int num = 0, pre = b[1].y;
  long long ans = 1;
  for (int i = 1; i <= n; i++) {
    if (b[i].y == pre)
      num++;
    else {
      pre = b[i].y;
      ans = (ans * sui[num]) % mod;
      num = 1;
    }
  }
  ans = (ans * sui[num]) % mod;
  return ans;
}
int main() {
  int t;
  cin >> n;
  long long cnt = 1;
  for (int i = 1; i <= n; i++) {
    cnt = (cnt * i) % mod;
    sui[i] = cnt;
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[i].x >> a[i].y;
    a[i].id = i;
    b[i] = a[i];
  }
  sort(a + 1, a + 1 + n, cmp1);
  sort(b + 1, b + 1 + n, cmp2);
  long long ans = sui[n] - slove1() - slove2() + judge();
  ans = (ans + mod * 3) % mod;
  cout << ans << endl;
}
