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 = 200001;
  const int mod = 998244353;
  int n, m;
  int sum[maxn << 2];
  int y[maxn], x;
  int vis[maxn];
  int fac[maxn << 1], rfac[maxn << 1];
  int mul(int a, int b) { return 1LL * a * b % mod; }
  int qpow(int a, int q) {
    int res = 1;
    while (q) {
      if (q & 1) res = mul(res, a);
      q >>= 1;
      a = mul(a, a);
    }
    return res;
  }
  void init() {
    fac[0] = 1;
    int n = maxn * 2;
    for (int i = 1; i < n; ++i) {
      fac[i] = mul(fac[i - 1], i);
    }
    rfac[n - 1] = qpow(fac[n - 1], mod - 2);
    for (int i = n - 2; i >= 0; --i) {
      rfac[i] = mul(rfac[i + 1], i + 1);
    }
  }
  void pushup(int rt) { sum[rt] = sum[(rt << 1)] + sum[(rt << 1 | 1)]; }
  void build(int rt, int l, int r) {
    if (l == r) {
      sum[rt] = 1;
      return;
    }
    int mid = (l + r) / 2;
    build((rt << 1), l, mid);
    build((rt << 1 | 1), mid + 1, r);
    pushup(rt);
  }
  int query(int rt, int l, int r, int val) {
    if (l == r) {
      return l;
    }
    int mid = (l + r) >> 1;
    if (val <= sum[(rt << 1)]) return query((rt << 1), l, mid, val);
    return query((rt << 1 | 1), mid + 1, r, val - sum[(rt << 1)]);
  }
  void update(int rt, int l, int r, int pos, int val) {
    if (l == r) {
      sum[rt] = val;
      return;
    }
    int mid = (l + r) >> 1;
    if (pos <= mid)
      update((rt << 1), l, mid, pos, val);
    else
      update((rt << 1 | 1), mid + 1, r, pos, val);
    pushup(rt);
  }
  int C(int a, int b) { return mul(mul(fac[a], rfac[b]), rfac[a - b]); }
  int main() {
    int t;
    scanf("%d", &t);
    init();
    int len = maxn - 1;
    build(1, 1, len);
    while (t--) {
      scanf("%d%d", &n, &m);
      for (int i = 0; i < m; ++i) {
        scanf("%d%d", &x, &y[i]);
      }
      memset(vis, 0, sizeof(vis));
      int c = 0;
      vector<int> ve;
      for (int i = m - 1; i >= 0; --i) {
        int p = query(1, 1, len, y[i]);
        int q = query(1, 1, len, y[i] + 1);
        if (!vis[q]) {
          vis[q] = 1;
          ++c;
        }
        update(1, 1, len, p, 0);
        ve.push_back(p);
      }
      printf("%d\n", C(2 * n - c - 1, n));
      for (auto u : ve) {
        update(1, 1, len, u, 1);
      }
    }
  }
  