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;
long long int mod = 998244353;
long double eps = 1e-12;
long long int dx[] = {0, 0, 1, -1, 1, 1, -1, -1};
long long int dy[] = {1, -1, 0, 0, 1, -1, 1, -1};
long long int n, res = 0, k = 0, h, ans = 0, m, level = 0;
;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int x, long long int y) {
  return (x * y) / gcd(x, y);
}
long long int pot(long long int a, long long int b) {
  long long int r = 1;
  while (b) {
    if (b % 2) r = (r * a) % mod;
    a = (a * a) % mod;
    b /= 2;
  }
  return r % mod;
}
void pr(long long int a[], long long int n) {
  for (long long int i = 1; i <= n; i++) cout << a[i] << " ";
  cout << '\n';
}
void yes() { cout << "YES\n"; }
void no() { cout << "NO\n"; }
vector<long long int> v[1000005], vnode[1000005];
void add_edge(long long int x, long long int y) {
  v[x].push_back(y);
  v[y].push_back(x);
}
string s2, ch = "", s, s1 = "", t;
vector<pair<string, long long int> > vm;
vector<pair<pair<long long int, long long int>, long long int> > vvv, vvvr;
vector<pair<long long int, long long int> > vv[1000005], vv2;
long long int b[1000005] = {0};
long long int a[1000005], lps[1000005];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long int i = 0, x = 0, z = 0, y = 0, j = 0, q, mx = 0, mm, idx = 0,
                ok = 0, l = 0, r, negatif = 0, positif = 0, l1, r1, p;
  set<pair<long long int, long long int> >::iterator it, it2;
  multiset<long long int>::iterator it1;
  multiset<long long int> mst1, mst2;
  cin >> n;
  vector<pair<long long int, long long int> > vv1, vv2;
  res = 1;
  for (i = 1; i <= (long long int)(n); i++) {
    cin >> x >> y;
    vv1.push_back({x, y});
    vv2.push_back({y, x});
    res *= i;
    res %= mod;
  }
  sort(vv1.begin(), vv1.end());
  sort(vv2.begin(), vv2.end());
  long long int ans = 1;
  for (i = 0; i < n; i++) {
    long long int cnt = 1;
    long long int idx = 1;
    for (j = i; j < n; j++) {
      if (vv1[j].first == vv1[i].first) {
        cnt *= idx;
        idx++;
        cnt %= mod;
      } else
        break;
    }
    if (j > i) i = j - 1;
    ans *= cnt;
    ans %= mod;
  }
  res -= ans;
  res = (res % mod + mod) % mod;
  ans = 1;
  for (i = 0; i < n; i++) {
    long long int cnt = 1;
    long long int idx = 1;
    for (j = i; j < n; j++) {
      if (vv2[j].first == vv2[i].first) {
        cnt *= idx;
        idx++;
        cnt %= mod;
      } else
        break;
    }
    if (j > i) i = j - 1;
    ans *= cnt;
    ans %= mod;
  }
  res -= ans;
  res = (res % mod + mod) % mod;
  ans = 1;
  bool sorted = 1;
  for (i = 0; i < n; i++) {
    long long int cnt = 1;
    long long int idx = 1;
    for (j = i; j < n; j++) {
      if (vv1[j].first == vv1[i].first && vv1[j].second == vv1[i].second) {
        cnt *= idx;
        idx++;
        cnt %= mod;
      } else
        break;
    }
    if (j > i) i = j - 1;
    ans *= cnt;
    ans %= mod;
  }
  for (i = 1; i < n; i++) {
    if (vv1[i].second < vv1[i - 1].second) sorted = 0;
  }
  if (sorted) res += ans;
  res = (res % mod + mod) % mod;
  cout << res;
  return 0;
}
