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 N = 400005, mod = 998244353;
  int qp(int a, int b) {
    int ans = 1;
    while (b) {
      if (b & 1) ans = 1ll * ans * a % mod;
      a = 1ll * a * a % mod;
      b >>= 1;
    }
    return ans;
  }
  int n, m, fc[N], iv[N], x[N], y[N], t, b[N], vis[N];
  int c(int n, int m) { return 1ll * fc[n] * iv[m] % mod * iv[n - m] % mod; }
  vector<int> v;
  void add(int x, int v) {
    for (int i = x; i <= n; i += i & (-i)) b[i] += v;
  }
  int ask(int x) {
    int v = 0;
    for (int i = x; i; i -= i & (-i)) v += b[i];
    return v;
  }
  void sol() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; i++) scanf("%d%d", &x[i], &y[i]);
    t = 0, v.clear();
    for (int i = m; i >= 1; i--) {
      int l = y[i], r = n, p = 1e9;
      while (l <= r) {
        int mid = l + r >> 1;
        int f = ask(mid);
        if (mid - f < y[i]) {
          l = mid + 1;
          continue;
        }
        if (mid - f > y[i]) {
          r = mid - 1;
          continue;
        }
        p = min(p, mid);
        r = mid - 1;
      }
      if (!vis[p + 1]) t++;
      add(p, 1);
      v.push_back(p);
      vis[p] = 1;
    }
    printf("%d\n", c(n + n - 1 - t, n));
    for (auto i : v) add(i, -1), vis[i] = 0;
  }
  int main() {
    srand((unsigned long long)new char);
    fc[0] = iv[0] = 1;
    for (int i = 1; i <= 400000; i++)
      fc[i] = 1ll * fc[i - 1] * i % mod, iv[i] = qp(fc[i], mod - 2);
    int t;
    scanf("%d", &t);
    while (t--) sol();
    return 0;
  }
  