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;
int main() {
  long long MOD = 998244353;
  long long n;
  cin >> n;
  vector<pair<long long, long long>> v;
  vector<long long> fir;
  vector<long long> sec;
  long long x1, x2;
  for (int i = 0; i < n; ++i) {
    cin >> x1 >> x2;
    fir.push_back(x1);
    sec.push_back(x2);
    v.push_back(make_pair(x1, x2));
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < v.size(); ++i) {
  }
  long long prev = -1;
  vector<long long> dp;
  dp.push_back(1);
  long long curCnt = 1;
  long long curVal = v[0].second;
  long long curBig = v[0].second;
  for (int i = 1; i < n; ++i) {
    if (v[i - 1].first != v[i].first) {
      if (v[i - 1].second <= v[i].second) {
        dp.push_back(dp[i - 1]);
        prev = curBig;
        curCnt = 1;
        curVal = v[i].second;
        curBig = v[i].second;
      } else {
        dp.push_back(0);
        break;
      }
      continue;
    }
    if (curBig < v[i].second) {
      dp.push_back(dp[i - 1]);
      curCnt = 1;
      curVal = v[i].second;
      if (curBig < v[i].second) curBig = v[i].second;
      continue;
    }
    long long cnt = 0;
    if (v[i].second == curVal) {
      cnt = curCnt;
      curCnt++;
    }
    if (cnt == 0) {
      if (prev > v[i].second) {
        dp.push_back(0);
        break;
      }
      dp.push_back(dp[i - 1]);
      curCnt = 1;
      curVal = v[i].second;
      if (curBig < v[i].second) curBig = v[i].second;
      continue;
    }
    dp.push_back(dp[i - 1] * (cnt + 1) % MOD);
    if (curBig < v[i].second) curBig = v[i].second;
  }
  sort(fir.begin(), fir.end());
  sort(sec.begin(), sec.end());
  long long lastCnt = 1;
  long long ans1 = 1;
  long long ans2 = 1;
  for (int i = 1; i < fir.size(); ++i) {
    if (fir[i] != fir[i - 1]) {
      lastCnt = 1;
    } else {
      ans1 = (ans1 * (lastCnt + 1)) % MOD;
      lastCnt++;
    }
  }
  lastCnt = 1;
  for (int i = 1; i < sec.size(); ++i) {
    if (sec[i] != sec[i - 1]) {
      lastCnt = 1;
    } else {
      ans2 = (ans2 * (lastCnt + 1)) % MOD;
      lastCnt++;
    }
  }
  long long all = 1;
  for (int i = 0; i < n; ++i) {
    all = (all * (i + 1)) % MOD;
  }
  cout << (all - ans1 - ans2 + dp[dp.size() - 1] + MOD * 2) % MOD << endl;
  return 0;
}
