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;
  const int M = 998244353;
  int add(int a, int b) {
    a += b;
    if (a >= M) return a - M;
    if (a < 0) return a + M;
    return a;
  }
  int mul(int a, int b) { return a * (long long)b % M; }
  int pw(int a, int b) {
    int r = 1;
    while (b) {
      if (b & 1) r = mul(r, a);
      a = mul(a, a);
      b /= 2;
    }
    return r;
  }
  int dv(int a, int b) { return mul(a, pw(b, M - 2)); }
  const int N = (int)2e5 + 7;
  int n, m, x[N], y[N], fact[2 * N], ifact[2 * N], aib[N];
  int comb(int n, int k) { return mul(fact[n], mul(ifact[k], ifact[n - k])); }
  void addAib(int i, int x) {
    while (i < N) {
      aib[i] += x;
      i += i & (-i);
    }
  }
  int getAib(int i) {
    int sol = 0;
    while (i > 0) {
      sol += aib[i];
      i -= i & (-i);
    }
    return sol;
  }
  bool luat[N];
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    fact[0] = 1;
    for (int i = 1; i < 2 * N; i++) {
      fact[i] = mul(fact[i - 1], i);
    }
    ifact[2 * N - 1] = dv(1, fact[2 * N - 1]);
    for (int i = 2 * N - 2; i >= 0; i--) {
      ifact[i] = mul(ifact[i + 1], i + 1);
    }
    for (int i = 1; i < N; i++) {
      addAib(i, +1);
    }
    int t;
    cin >> t;
    while (t--) {
      cin >> n >> m;
      for (int i = 1; i <= m; i++) {
        cin >> x[i] >> y[i];
      }
      int cnt = 0;
      vector<int> kek;
      for (int j = m; j >= 1; j--) {
        int i = x[j], what = y[j];
        int sol = 0, step = (1 << 20), cur = 0;
        while (step) {
          if (sol + step < N && cur + aib[sol + step] < what) {
            sol += step;
            cur += aib[sol];
          }
          step /= 2;
        }
        cnt += luat[sol];
        luat[sol + 1] = 1;
        addAib(sol + 1, -1);
        kek.push_back(sol + 1);
      }
      for (auto &i : kek) {
        if (i < n) {
          cnt += 1 - luat[i + 1];
        }
      }
      for (auto &i : kek) {
        addAib(i, +1);
        luat[i] = 0;
      }
      cout << comb(2 * n - 1 - cnt, n) << "\n";
    }
    return 0;
  }
  