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 N = 4 * 1e5 + 100;
  long long t, n, m, x[N], y[N], fac[N], inv[N], pos[N];
  inline void add(long long &a, long long b) {
    a = ((a + b >= 998244353) ? a + b - 998244353 : a + b);
  }
  inline void del(long long &a, long long b) {
    a = ((a - b < 0) ? a - b + 998244353 : a - b);
  }
  inline void mul(long long &a, long long b) { a = a * b % 998244353; }
  inline long long C(long long n, long long m) {
    return fac[n] * inv[m] % 998244353 * inv[n - m] % 998244353;
  }
  inline long long m_pow(long long a, long long b) {
    long long ans = 1;
    while (b) {
      if (b & 1) mul(ans, a);
      b >>= 1;
      mul(a, a);
    }
    return ans;
  }
  inline long long read() {
    long long f = 1, x = 0;
    char s = getchar();
    while (s < '0' || s > '9') {
      if (s == '-') f = -1;
      s = getchar();
    }
    while (s >= '0' && s <= '9') {
      x = x * 10 + s - '0';
      s = getchar();
    }
    return x * f;
  }
  struct bit {
    long long sh[N], ti[N];
    inline long long lowbit(long long x) { return x & (-x); }
    inline void change(long long x, long long v) {
      while (x <= n) sh[x] += v, x += lowbit(x);
    }
    inline long long query(long long x) {
      long long ans = 0;
      while (x) ans += sh[x], x -= lowbit(x);
      return ans;
    }
    inline long long kth(long long k) {
      long long x = 0, sum = 0, lg = log(n) / log(2);
      for (long long i = lg; i >= 0; i--) {
        if (sh[x + (1 << i)] < k) {
          k -= sh[x + (1 << i)];
          x += (1 << i);
        }
      }
      return x + 1;
    }
  } T;
  void prework() {
    n = 400000;
    fac[0] = 1;
    for (long long i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % 998244353;
    inv[n] = m_pow(fac[n], 998244353 - 2);
    for (long long i = n - 1; i >= 0; i--)
      inv[i] = inv[i + 1] * (i + 1) % 998244353;
    for (long long i = 1; i <= n / 2; i++) T.change(i, 1);
  }
  signed main() {
    prework();
    t = read();
    while (t--) {
      n = read();
      m = read();
      for (long long i = 1; i <= m; i++) x[i] = read(), y[i] = read();
      long long k = 0;
      for (long long i = m; i >= 1; i--) {
        pos[i] = T.kth(y[i]);
        if (T.query(pos[i] + 1) > y[i]) k++;
        T.change(pos[i], -1);
      }
      printf("%lld\n", C(2 * n - k - 1, n));
      for (long long i = 1; i <= m; i++) T.change(pos[i], 1);
    }
  }
  