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 = 600010;
  const int mod = 998244353;
  long long mpow(long long a, long long b) {
    long long ans = 1;
    while (b) {
      if (b & 1) ans = ans * a % mod;
      b >>= 1;
      a = a * a % mod;
    }
    return ans;
  }
  int fac[N], inv[N];
  int com(int n, int m) {
    if (n < m) return 0;
    return (long long)fac[n] * inv[m] % mod * inv[n - m] % mod;
  }
  int c[N], a[N], b[N];
  int lowbit(int x) { return x & (-x); }
  void insert(int &x, int y) {
    for (int i = x; i < N; i += lowbit(i)) c[i] += y;
  }
  int query(int x) {
    int sum = 0, pos = 0;
    for (int i = 17; i >= 0; i--) {
      if ((pos | (1 << i)) >= N) continue;
      if (sum + c[pos | (1 << i)] >= x) continue;
      pos |= (1 << i);
      sum += c[pos];
    }
    return pos + 1;
  }
  void solve() {
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; i++) scanf("%d%d", &a[i], &b[i]);
    set<int> st;
    stack<int> stk;
    for (int i = m; i >= 1; i--) {
      int pos = query(b[i]);
      stk.push(pos);
      insert(pos, -1);
      pos = query(b[i]);
      st.insert(pos);
    }
    int tot = st.size();
    int ans = com(2 * n - tot - 1, n);
    printf("%d\n", ans);
    while (!stk.empty()) {
      int x = stk.top();
      stk.pop();
      insert(x, 1);
    }
  }
  int main() {
    fac[0] = inv[0] = 1;
    for (int i = 1; i < N; i++) {
      fac[i] = fac[i - 1] * (long long)i % mod;
      inv[i] = mpow(fac[i], mod - 2);
    }
    for (int i = 1; i < N; i++) insert(i, 1);
    int t;
    scanf("%d", &t);
    while (t--) solve();
    return 0;
  }
  