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>
  #pragma GCC optimize("Ofast")
  #pragma loop - opt(on)
  using namespace std;
  struct BIT {
    static const long long int n = (1 << 19), lg = 19;
    vector<long long int> cnt = vector<long long int>(n + 1, 0);
    stack<long long int> op;
    long long int lowbit(long long int x) { return x & (-x); }
    void modify(long long int x, long long int v) {
      if (v < 0) op.push(x);
      for (long long int i = x; i <= n; i += lowbit(i)) {
        cnt[i] += v;
      }
    }
    long long int query(long long int k) {
      long long int ans = 0, cal = 0;
      for (long long int i = lg - 1; i >= 0; i--)
        if (cal + cnt[ans | (1 << i)] < k) {
          ans = (ans | (1 << i)), cal += cnt[ans];
        }
      return ans + 1;
    }
    void init_() {
      while (op.size()) {
        long long int x = op.top();
        op.pop();
        modify(x, 1);
      }
    }
  };
  namespace solver {
  long long int n, m;
  BIT bit;
  const long long int P = (1 << 19);
  vector<long long int> x, y, fac, inv;
  vector<pair<long long int, long long int> > a;
  long long int pow_(long long int a, long long int times) {
    long long int ans = 1;
    for (; times > 0; times >>= 1, a = a * a % 998244353) {
      if (times & 1) ans = ans * a % 998244353;
    }
    return ans;
  }
  void build() {
    fac.assign(P, 1);
    inv.assign(P, 1);
    for (long long int i = 1; i <= P - 1; i++)
      fac[i] = fac[i - 1] * i % 998244353;
    inv[P - 1] = pow_(fac[P - 1], 998244353 - 2);
    for (long long int i = P - 2; i >= 0; i--)
      inv[i] = inv[i + 1] * (i + 1) % 998244353;
    for (long long int i = 1; i <= P; i++) bit.modify(i, 1);
  }
  long long int C(long long int n, long long int m) {
    return fac[n] * inv[m] % 998244353 * inv[n - m] % 998244353;
  }
  void init_(long long int _n, long long int _m) {
    n = _n, m = _m;
    x.assign(m + 1, 0);
    y.assign(m + 1, 0);
    a.clear();
    bit.init_();
  }
  long long int solve() {
    for (long long int i = m; i >= 1; i--) {
      long long int id = bit.query(y[i]);
      a.push_back({id, x[i]});
      bit.modify(id, -1);
    }
    long long int c = m;
    sort(a.begin(), a.end());
    for (long long int i = 1; i <= m - 1; i++) {
      if (a[i].first == a[i - 1].first + 1 && a[i].second > a[i - 1].second) c--;
    }
    return C(2 * n - 1 - c, n);
  }
  };  // namespace solver
  using namespace solver;
  signed main() {
    ios::sync_with_stdio(false), cin.tie(0);
    long long int t;
    cin >> t;
    build();
    while (t--) {
      long long int n, m;
      cin >> n >> m;
      init_(n, m);
      for (long long int i = 1; i <= m; i++) cin >> x[i] >> y[i];
      cout << solve() << "\n";
    }
    return 0;
  }
  