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 maxn = 2e5 + 5;
  const int maxm = 2e5;
  const int mod = 998244353;
  int x[maxn], y[maxn], fac[maxn << 1], fav[maxn << 1];
  int t[maxn << 2], vis[maxn];
  void pushup(int rt) { t[rt] = t[rt << 1] + t[rt << 1 | 1]; }
  void build(int rt, int l, int r) {
    if (l == r) {
      t[rt]++;
      return;
    }
    int mid = l + r >> 1;
    build(rt << 1, l, mid);
    build(rt << 1 | 1, mid + 1, r);
    pushup(rt);
  }
  void add(int rt, int l, int r, int x, int v) {
    if (l == r) {
      t[rt] += v;
      return;
    }
    int mid = l + r >> 1;
    if (x <= mid) add(rt << 1, l, mid, x, v);
    if (x >= mid + 1) add(rt << 1 | 1, mid + 1, r, x, v);
    pushup(rt);
  }
  int qry(int rt, int l, int r, int k) {
    if (l == r) return l;
    int mid = l + r >> 1;
    if (t[rt << 1] >= k) return qry(rt << 1, l, mid, k);
    return qry(rt << 1 | 1, mid + 1, r, k - t[rt << 1]);
  }
  int qpow(int a, int b, int c) {
    int ans = 1;
    while (b) {
      if (b & 1) ans = 1ll * ans * a % c;
      b >>= 1;
      a = 1ll * a * a % c;
    }
    return ans;
  }
  int C(int a, int b) {
    if (a < b) return 0;
    return 1ll * fac[a] * fav[a - b] % mod * fav[b] % mod;
  }
  int main() {
    fac[0] = 1;
    for (int i = 1; i < (maxn << 1); ++i) fac[i] = 1ll * fac[i - 1] * i % mod;
    fav[(maxn << 1) - 1] = qpow(fac[(maxn << 1) - 1], mod - 2, mod);
    for (int i = (maxn << 1) - 2; i >= 0; --i)
      fav[i] = 1ll * fav[i + 1] * (i + 1) % mod;
    build(1, 1, maxm);
    int t;
    scanf("%d", &t);
    while (t--) {
      int n, m;
      scanf("%d%d", &n, &m);
      for (int i = 0; i < m; ++i) {
        scanf("%d%d", &x[i], &y[i]);
      }
      int cnt = 0;
      vector<int> now, vs;
      for (int i = m - 1; i >= 0; --i) {
        int u = qry(1, 1, maxm, y[i]);
        if (!vis[u]) cnt++;
        vis[u] = 1;
        vs.push_back(u);
        now.push_back(qry(1, 1, maxm, y[i] - 1));
        add(1, 1, maxm, now.back(), -1);
      }
      printf("%d\n", C(2 * n - 1 - cnt, n));
      for (auto u : now) add(1, 1, maxm, u, 1);
      for (auto u : vs) vis[u] = 0;
    }
  }
  