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 long long mod = 998244353;
  int n, m, y[200011], p[200011], id;
  int v[200011];
  struct sgt {
    int sum[800011];
    void clear(int k, int l, int r) {
      sum[k] = r - l + 1;
      if (l == r) return;
      int mid = (l + r) >> 1;
      clear(k << 1, l, mid);
      clear(k << 1 | 1, mid + 1, r);
    }
    void modify(int k, int l, int r, int w, int num) {
      if (l == r) {
        sum[k] = num;
        return;
      }
      int mid = (l + r) >> 1;
      if (w <= mid)
        modify(k << 1, l, mid, w, num);
      else
        modify(k << 1 | 1, mid + 1, r, w, num);
      sum[k] = sum[k << 1] + sum[k << 1 | 1];
    }
    int find(int k, int l, int r, int rk) {
      if (l == r) return l;
      int mid = (l + r) >> 1;
      if (mid - l + 1 - sum[k << 1] >= rk)
        return find(k << 1, l, mid, rk);
      else
        return find(k << 1 | 1, mid + 1, r, rk - (mid - l + 1 - sum[k << 1]));
    }
  } t;
  long long inp[400011];
  long long pw(long long s1, long long s2) {
    if (!s2) return 1;
    if (s2 % 2) return s1 * pw(s1, s2 - 1) % mod;
    long long ret = pw(s1, s2 / 2);
    return ret * ret % mod;
  }
  void solve() {
    id++;
    int c = 0;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; i++) scanf("%d%d", &y[i], &y[i]);
    for (int i = m; i >= 1; i--) {
      int l = t.find(1, 1, n, y[i]), r = t.find(1, 1, n, y[i] + 1);
      if (v[r] != id) c++;
      v[r] = id;
      t.modify(1, 1, n, l, 1);
      p[i] = l;
    }
    for (int i = 1; i <= m; i++) t.modify(1, 1, n, p[i], 0);
    cout << inp[2 * n - 1 - c] * pw(inp[n], mod - 2) % mod *
                pw(inp[n - 1 - c], mod - 2) % mod
         << endl;
  }
  int main() {
    inp[0] = 1;
    for (int i = 1; i <= 400000; i++) inp[i] = inp[i - 1] * i % mod;
    int t;
    scanf("%d", &t);
    while (t--) solve();
    return 0;
  }
  