description: Consider the insertion sort algorithm used to sort an integer sequence [a_1, a_2, …, a_n] of length n in non-decreasing order.
  
  For each i in order from 2 to n, do the following. If a_i ≥ a_{i-1}, do nothing and move on to the next value of i. Otherwise, find the smallest j such that a_i < a_j, shift the elements on positions from j to i-1 by one position to the right, and write down the initial value of a_i to position j. In this case we'll say that we performed an insertion of an element from position i to position j.
  
  It can be noticed that after processing any i, the prefix of the sequence [a_1, a_2, …, a_i] is sorted in non-decreasing order, therefore, the algorithm indeed sorts any sequence.
  
  For example, sorting [4, 5, 3, 1, 3] proceeds as follows: 
  
    * i = 2: a_2 ≥ a_1, do nothing; 
    * i = 3: j = 1, insert from position 3 to position 1: [3, 4, 5, 1, 3]; 
    * i = 4: j = 1, insert from position 4 to position 1: [1, 3, 4, 5, 3]; 
    * i = 5: j = 3, insert from position 5 to position 3: [1, 3, 3, 4, 5]. 
  
  
  
  You are given an integer n and a list of m integer pairs (x_i, y_i). We are interested in sequences such that if you sort them using the above algorithm, exactly m insertions will be performed: first from position x_1 to position y_1, then from position x_2 to position y_2, ..., finally, from position x_m to position y_m.
  
  How many sequences of length n consisting of (not necessarily distinct) integers between 1 and n, inclusive, satisfy the above condition? Print this number modulo 998 244 353.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^5). Description of the test cases follows.
  
  The first line of each test case contains two integers n and m (2 ≤ n ≤ 2 ⋅ 10^5; 0 ≤ m < n) — the length of the sequence and the number of insertions.
  
  The i-th of the following m lines contains two integers x_i and y_i (2 ≤ x_1 < x_2 < … < x_m ≤ n; 1 ≤ y_i < x_i). These lines describe the sequence of insertions in chronological order.
  
  It is guaranteed that the sum of m over all test cases does not exceed 2 ⋅ 10^5. Note that there is no constraint on the sum of n of the same kind.
  
  Output
  
  For each test case, print the number of sequences of length n consisting of integers from 1 to n such that sorting them with the described algorithm produces the given sequence of insertions, modulo 998 244 353.
  
  Example
  
  Input
  
  
  3
  3 0
  3 2
  2 1
  3 1
  5 3
  3 1
  4 1
  5 3
  
  
  Output
  
  
  10
  1
  21
  
  Note
  
  In the first test case, the algorithm performs no insertions — therefore, the initial sequence is already sorted in non-decreasing order. There are 10 such sequences: [1, 1, 1], [1, 1, 2], [1, 1, 3], [1, 2, 2], [1, 2, 3], [1, 3, 3], [2, 2, 2], [2, 2, 3], [2, 3, 3], [3, 3, 3].
  
  In the second test case, the only sequence satisfying the conditions is [3, 2, 1].
  
  In the third test case, [4, 5, 3, 1, 3] is one of the sought sequences.
solution: #include <bits/stdc++.h>
  using namespace std;
  using ll = long long;
  const ll MOD = 998244353;
  const int MX = 2e5;
  struct {
    int st[MX + 5];
    void upd(int t, int v) {
      while (t <= MX) st[t] += v, t += t & -t;
    }
    int get(int t) {
      int r = 0;
      while (t) r += st[t], t -= t & -t;
      return r;
    }
  } fw;
  ll F[MX * 2], F_[MX * 2];
  void sol() {
    int n, m;
    cin >> n >> m;
    vector<int> x, y;
    for (int i = 0, u, v; i < m; i++)
      cin >> u >> v, x.emplace_back(u), y.emplace_back(v);
    vector<pair<int, int>> p;
    int k = m;
    for (int i = m - 1; i >= 0; i--) {
      int s = y[i], e = n;
      while (s < e) {
        int m = (s + e) / 2;
        if (m - fw.get(m) >= y[i])
          e = m;
        else
          s = m + 1;
      }
      p.emplace_back(s, x[i]);
      fw.upd(s, 1);
    }
    sort(p.begin(), p.end());
    for (int i = 0; i < m - 1; i++)
      if (p[i].first + 1 == p[i + 1].first && p[i].second < p[i + 1].second) k--;
    for (auto &i : p) fw.upd(i.first, -1);
    cout << F[2 * n - k - 1] * F_[n] % MOD * F_[n - k - 1] % MOD << '\n';
  }
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    F[0] = 1;
    for (int i = 1; i < 2 * MX; i++) F[i] = F[i - 1] * i % MOD;
    ll &x = F_[2 * MX - 1];
    x = 1;
    for (int i = 30; i >= 0; i--) {
      x = x * x % MOD;
      if ((1 << i) & (MOD - 2)) x = x * F[2 * MX - 1] % MOD;
    }
    for (int i = 2 * MX - 2; i >= 0; i--) F_[i] = F_[i + 1] * (i + 1) % MOD;
    int t;
    cin >> t;
    while (t--) sol();
    return 0;
  }
  