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 N = 1 << 19;
  const int MO = 998244353;
  int mul(int x, int y) { return (long long)x * y % MO; }
  int fpow(int x, int y = MO - 2) {
    int z = 1;
    while (y) {
      if (y & 1) z = mul(z, x);
      x = mul(x, x);
      y >>= 1;
    }
    return z;
  }
  int n, m, f[N], g[N], x[N], y[N], z[N], a[N], t[N];
  void modify(int x, int k = 1, int l = 1, int r = n) {
    t[k]++;
    if (l == r) return;
    int h = l + r >> 1;
    if (x <= h)
      modify(x, k << 1, l, h);
    else
      modify(x, k << 1 | 1, h + 1, r);
  }
  int query(int x, int k = 1, int l = 1, int r = n) {
    if (l == r) return l;
    int h = l + r >> 1;
    int o = h - l + 1 - t[k << 1];
    if (x <= o) return query(x, k << 1, l, h);
    return query(x - o, k << 1 | 1, h + 1, r);
  }
  void clr(int k = 1) {
    if (k >= N || !t[k]) return;
    t[k] = 0;
    clr(k << 1);
    clr(k << 1 | 1);
  }
  int main() {
    int T, i, o;
    n = N - 1;
    f[0] = 1;
    for (i = 1; i <= n; i++) f[i] = mul(f[i - 1], i);
    g[n] = fpow(f[n]);
    for (i = n; i >= 1; i--) g[i - 1] = mul(g[i], i);
    scanf("%d", &T);
    while (T--) {
      scanf("%d%d", &n, &m);
      for (i = 1; i <= m; i++) scanf("%d%d", x + i, y + i);
      for (i = m; i >= 1; i--) {
        z[i] = query(y[i]);
        a[z[i]] = x[i];
        modify(z[i]);
      }
      o = m;
      for (i = 1; i <= m; i++)
        if (a[z[i] + 1] > x[i]) o--;
      o = n - 1 - o;
      printf("%d\n", mul(f[n + o], mul(g[n], g[o])));
      clr();
      for (i = 1; i <= m; i++) a[z[i]] = 0;
    }
    return 0;
  }
  