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;
  using ll = long long;
  constexpr ll mod = 998244353;
  constexpr int N = 400000;
  ll fac[N + 1], ifac[N + 1];
  ll qpow(ll x, ll y, ll p = mod) {
    ll res = 1;
    x %= p;
    while (y) {
      if (y & 1) res = res * x % p;
      x = x * x % p;
      y >>= 1;
    }
    return res;
  }
  ll inv(ll x, ll p = mod) { return qpow(x, p - 2, p); }
  ll comb(int x, int y) {
    assert(x >= y);
    return fac[x] * ifac[x - y] % mod * ifac[y] % mod;
  }
  int t[N];
  void add(int x, int v) {
    while (x <= 200000) {
      t[x] += v;
      x += x & -x;
    }
  }
  int kth(int k) {
    int cnt = 0, num = 0;
    for (int i = 20; i >= 0; i--) {
      num += 1 << i;
      if (num >= 200000 || cnt + t[num] >= k) {
        num -= 1 << i;
      } else {
        cnt += t[num];
      }
    }
    return num + 1;
  }
  int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    cout << fixed << setprecision(20);
    fac[0] = 1;
    for (int i = 1; i <= N; i++) fac[i] = fac[i - 1] * i % mod;
    ifac[N] = inv(fac[N]);
    for (int i = N; i > 0; i--) ifac[i - 1] = ifac[i] * i % mod;
    for (int i = 1; i <= 200000; i++) add(i, 1);
    int T;
    cin >> T;
    while (T--) {
      int n, m;
      cin >> n >> m;
      vector<int> x(m), y(m);
      for (int i = 0; i < m; i++) {
        cin >> x[i] >> y[i];
      }
      set<int> S;
      for (int i = m - 1; i >= 0; i--) {
        S.insert(kth(y[i] + 1));
        y[i] = kth(y[i]);
        add(y[i], -1);
      }
      for (int i = 0; i < m; i++) {
        add(y[i], 1);
      }
      ll ans = comb(2 * n - 1 - S.size(), n);
      cout << ans << '\n';
    }
    return 0;
  }
  