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 mod = 998244353;
  int qpow(int a, int b) {
    int res = 1;
    while (b) {
      if (b & 1) res = (long long)res * a % mod;
      a = (long long)a * a % mod;
      b >>= 1;
    }
    return res;
  }
  int fac[200015 << 1], ifac[200015 << 1];
  int C(int n, int m) {
    if (n < m || m < 0) return 0;
    return (long long)fac[n] * ifac[m] % mod * ifac[n - m] % mod;
  }
  int t, n, m, x[200015], y[200015];
  int sum[200015];
  void add(int p, int v) {
    for (; p <= n; p += ((p) & (-p))) sum[p] += v;
  }
  int query(int p) {
    int res = 0;
    for (; p; p -= ((p) & (-p))) res += sum[p];
    return res;
  }
  int getpos(int x) {
    int l = 1, r = n, ans = 0;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (mid - query(mid) == x) {
        ans = mid;
      }
      if (mid - query(mid) < x)
        l = mid + 1;
      else
        r = mid - 1;
    }
    return ans;
  }
  int main() {
    fac[0] = 1;
    for (int i = (1); i <= (400000); i++)
      fac[i] = (long long)fac[i - 1] * i % mod;
    for (int i = (0); i <= (400000); i++) ifac[i] = qpow(fac[i], mod - 2);
    scanf("%d", &t);
    while (t--) {
      scanf("%d%d", &n, &m);
      for (int i = (1); i <= (m); i++) scanf("%d%d", &x[i], &y[i]);
      int cnt = n - 1;
      vector<int> vec;
      for (int i = (m); i >= (1); i--) {
        if (y[i] < n) {
          int p1 = getpos(y[i]);
          int p2 = getpos(y[i] + 1);
          if (p1 + 1 == p2) cnt--;
        }
        int p = getpos(y[i]);
        add(p, 1);
        vec.push_back(p);
      }
      printf("%d\n", C(n + cnt, n));
      for (auto x : vec) add(x, -1);
    }
    return 0;
  }
  