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;
  void read(int& x) {
    static char c;
    int f = 0;
    while (!isdigit(c = getchar()))
      if (c == '-') f = 1;
    x = c ^ 48;
    while (isdigit(c = getchar())) x = (x * 10) + (c ^ 48);
    if (f) x = -x;
  }
  void read(long long& x) {
    static char c;
    int f = 0;
    while (!isdigit(c = getchar()))
      if (c == '-') f = 1;
    x = c ^ 48;
    while (isdigit(c = getchar())) x = (x * 10) + (c ^ 48);
    if (f) x = -x;
  }
  void read(unsigned int& x) {
    static char c;
    while (!isdigit(c = getchar()))
      ;
    x = c ^ 48;
    while (isdigit(c = getchar())) x = (x * 10) + (c ^ 48);
  }
  void read(unsigned long long& x) {
    static char c;
    while (!isdigit(c = getchar()))
      ;
    x = c ^ 48;
    while (isdigit(c = getchar())) x = (x * 10) + (c ^ 48);
  }
  void chkmax(int& x, int y) {
    if (y > x) x = y;
  }
  void chkmin(int& x, int y) {
    if (y < x) x = y;
  }
  void chkmax(long long& x, long long y) {
    if (y > x) x = y;
  }
  void chkmin(long long& x, long long y) {
    if (y < x) x = y;
  }
  void chkmax(unsigned int& x, unsigned int y) {
    if (y > x) x = y;
  }
  void chkmin(unsigned int& x, unsigned int y) {
    if (y < x) x = y;
  }
  void chkmax(unsigned long long& x, unsigned long long y) {
    if (y > x) x = y;
  }
  void chkmin(unsigned long long& x, unsigned long long y) {
    if (y < x) x = y;
  }
  const int Mod = 998244353;
  void Add(int& x, int y) { (x += y) >= Mod ? x -= Mod : x; }
  int Pow(int a, int b) {
    int res = 1;
    while (b) {
      if (b & 1) res = 1ll * res * a % Mod;
      a = 1ll * a * a % Mod, b >>= 1;
    }
    return res;
  }
  const int N = 2e5 + 10;
  int n, m;
  int Fac[N << 1], Iac[N << 1];
  void Init() {
    const int _ = 4e5;
    Fac[0] = Iac[0] = 1;
    for (int i = 1; i <= _; ++i) Fac[i] = 1ll * Fac[i - 1] * i % Mod;
    Iac[_] = Pow(Fac[_], Mod - 2);
    for (int i = _ - 1; i; --i) Iac[i] = 1ll * Iac[i + 1] * (i + 1) % Mod;
  }
  int Binom(int a, int b) {
    return 1ll * Fac[a] * Iac[b] % Mod * Iac[a - b] % Mod;
  }
  namespace SPLAY {
  int tot, rt;
  struct Node {
    int fat, val, idx, siz, son[2];
    Node() {}
    Node(int _fat, int _val, int _idx)
        : fat(_fat), val(_val), idx(_idx), siz(_val) {
      son[0] = son[1] = 0;
    }
  } t[N << 1];
  void Splay_Init() { tot = rt = 0; }
  int NEW(int fat, int val, int idx) {
    t[++tot] = Node(fat, val, idx);
    return tot;
  }
  int GDir(int now) { return t[t[now].fat].son[1] == now; }
  void Update(int now) {
    t[now].siz = t[t[now].son[0]].siz + t[t[now].son[1]].siz + t[now].val;
  }
  void Rorate(int now) {
    int Fat = t[now].fat, Gat = t[Fat].fat;
    int Fdir = GDir(now), Gdir = GDir(Fat);
    if (Gat) t[Gat].son[Gdir] = now;
    t[now].fat = Gat, t[Fat].fat = now;
    t[Fat].son[Fdir] = t[now].son[Fdir ^ 1];
    t[t[now].son[Fdir ^ 1]].fat = Fat;
    t[now].son[Fdir ^ 1] = Fat;
    Update(Fat), Update(now);
  }
  void Splay(int now, int Goal = 0) {
    while (t[now].fat != Goal) {
      int Fat = t[now].fat, Gat = t[Fat].fat;
      if (Gat != Goal) (GDir(now) ^ GDir(Fat)) ? Rorate(now) : Rorate(Fat);
      Rorate(now);
    }
    if (!Goal) rt = now;
  }
  int Find(int pos) {
    int now = rt;
    while (true) {
      if ((pos >= t[t[now].son[0]].siz + 1) &&
          (pos <= t[t[now].son[0]].siz + t[now].val)) {
        int res = pos - t[t[now].son[0]].siz;
        return Splay(now), res;
      }
      if (pos <= t[t[now].son[0]].siz)
        now = t[now].son[0];
      else
        pos -= t[t[now].son[0]].siz + t[now].val, now = t[now].son[1];
    }
  }
  }  // namespace SPLAY
  using namespace SPLAY;
  void Solve() {
    read(n), read(m), Splay_Init();
    set<int> S;
    int lst = 0;
    for (int i = 1; i <= m; ++i) {
      int pos, ins;
      read(pos), read(ins);
      if (lst != pos) {
        if (!rt)
          rt = NEW(0, pos - lst - 1, lst + 1);
        else {
          int now = rt;
          while (t[now].son[1]) now = t[now].son[1];
          t[now].son[1] = NEW(now, pos - lst - 1, lst + 1), Splay(t[now].son[1]);
        }
      }
      int rep = Find(ins);
      if (rep > 1) {
        int nod = rt,
            _split = NEW(nod, t[nod].val - rep + 1, t[nod].idx + rep - 1);
        t[nod].val = rep - 1, S.insert(t[nod].idx + rep - 1);
        if (t[nod].son[1])
          t[t[nod].son[1]].fat = _split, t[_split].son[1] = t[nod].son[1];
        t[nod].son[1] = _split, t[_split].son[0] = NEW(_split, 1, pos);
        Splay(t[_split].son[0]);
      } else {
        int nod = rt;
        S.insert(t[nod].idx);
        if (!t[nod].son[0])
          t[nod].son[0] = NEW(nod, 1, pos), Splay(t[nod].son[0]);
        else {
          nod = t[nod].son[0];
          while (t[nod].son[1]) nod = t[nod].son[1];
          t[nod].son[1] = NEW(nod, 1, pos), Splay(t[nod].son[1]);
        }
      }
      lst = pos;
    }
    printf("%d\n", Binom(2 * n - 1 - S.size(), n));
  }
  int T;
  void Work() {
    read(T), Init();
    while (T--) Solve();
  }
  int main() { Work(); }
  