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 long long mod = 998244353;
  const int N = 1e6 + 9;
  int T, n, m, tot;
  struct pp {
    int x, y;
  } a[N];
  struct fhq {
    int rnd, ch[2], tag, val, mx, siz;
  } t[N];
  int New(int id) {
    ++tot;
    t[tot].ch[0] = t[tot].ch[1] = 0;
    t[tot].val = id, t[tot].mx = id, t[tot].tag = 0;
    t[tot].rnd = rand();
    t[tot].siz = 1;
    return tot;
  }
  void pushup(int c) {
    t[c].mx = max(t[c].val, max(t[t[c].ch[0]].mx, t[t[c].ch[1]].mx));
    t[c].siz = t[t[c].ch[0]].siz + t[t[c].ch[1]].siz + 1;
    return;
  }
  void upd(int c, int d) {
    if (!c) return;
    t[c].tag += d;
    t[c].val += d, t[c].mx += d;
    return;
  }
  void pushdown(int c) {
    if (!t[c].tag) return;
    upd(t[c].ch[0], t[c].tag);
    upd(t[c].ch[1], t[c].tag);
    t[c].tag = 0;
    return;
  }
  void split(int c, int k, int &x, int &y) {
    if (!c) {
      x = y = 0;
      return;
    }
    pushdown(c);
    if (t[c].val <= k) {
      x = c;
      split(t[c].ch[1], k, t[c].ch[1], y);
    } else {
      y = c;
      split(t[c].ch[0], k, x, t[c].ch[0]);
    }
    pushup(c);
    return;
  }
  int merge(int x, int y) {
    if (!x || !y) return x + y;
    if (t[x].rnd < t[y].rnd) {
      pushdown(x);
      t[x].ch[1] = merge(t[x].ch[1], y);
      pushup(x);
      return x;
    } else {
      pushdown(y);
      t[y].ch[0] = merge(x, t[y].ch[0]);
      pushup(y);
      return y;
    }
  }
  int rt;
  long long fac[N], inv[N];
  long long C(int x, int y) {
    if (x < 0 || y < 0 || y < x) return 0;
    return fac[y] * inv[x] % mod * inv[y - x] % mod;
  }
  int main() {
    srand(517);
    fac[0] = inv[0] = inv[1] = 1;
    for (int i = 1; i <= N - 9; i++) fac[i] = fac[i - 1] * i % mod;
    for (int i = 2; i <= N - 9; i++)
      inv[i] = inv[mod % i] * (mod - mod / i) % mod;
    for (int i = 1; i <= N - 9; i++) (inv[i] *= inv[i - 1]) %= mod;
    scanf("%d", &T);
    while (T--) {
      scanf("%d%d", &n, &m);
      for (int i = 1; i <= m; i++) {
        scanf("%d%d", &a[i].x, &a[i].y);
        a[i].x = i;
      }
      tot = 0, rt = 0;
      for (int i = 1; i <= m; i++) {
        int x, y, z;
        split(rt, a[i].y - 1, x, y);
        upd(y, 1);
        if (t[x].mx == a[i].y - 1 && t[x].mx) {
          split(x, a[i].y - 2, x, z);
        }
        rt = merge(merge(x, New(a[i].y)), y);
      }
      int x, y, z;
      int k = t[rt].siz + 1;
      long long ans = C(n, n * 2 - k);
      printf("%lld\n", ans);
    }
    return 0;
  }
  