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;
  struct Modint {
    long long val;
    Modint(long long _val = 0) : val(_val % MOD) {}
    Modint operator+(Modint other) const { return Modint(val + other.val); }
    void operator+=(Modint other) {
      val += other.val;
      val %= MOD;
    }
    Modint operator-() const { return Modint(MOD - val); }
    Modint operator-(Modint other) const { return Modint(val + MOD - other.val); }
    void operator-=(Modint other) {
      val += MOD - other.val;
      val %= MOD;
    }
    Modint operator*(Modint other) const { return Modint(val * other.val); }
    void operator*=(Modint other) {
      val *= other.val;
      val %= MOD;
    }
    bool operator==(Modint other) const { return val == other.val; }
    bool operator!=(Modint other) const { return val != other.val; }
  };
  Modint exp(Modint a, int k) {
    if (k == 0) {
      return Modint(1);
    } else if (k % 2 == 0) {
      Modint half = exp(a, k / 2);
      return half * half;
    } else {
      return a * exp(a, k - 1);
    }
  }
  Modint inv(Modint a) { return exp(a, MOD - 2); }
  ostream& operator<<(ostream& out, Modint p) {
    out << p.val;
    return out;
  }
  mt19937 randgen;
  struct Treap {
    struct Node {
      int value, lazy;
      unsigned int priority;
      Node* lch;
      Node* rch;
      Node(int new_value) {
        value = new_value;
        priority = randgen();
        lch = 0;
        rch = 0;
      }
      void propagate() {
        value += lazy;
        if (lch) {
          lch->lazy += lazy;
        }
        if (rch) {
          rch->lazy += lazy;
        }
        lazy = 0;
      }
    };
    Node* root = 0;
    pair<Node*, Node*> split(int key, Node* cur) {
      if (cur == 0) return {0, 0};
      cur->propagate();
      pair<Node*, Node*> result;
      if (key <= cur->value) {
        auto ret = split(key, cur->lch);
        cur->lch = ret.second;
        result = {ret.first, cur};
      } else {
        auto ret = split(key, cur->rch);
        cur->rch = ret.first;
        result = {cur, ret.second};
      }
      return result;
    }
    Node* merge(Node* left, Node* right) {
      if (left == 0) return right;
      if (right == 0) return left;
      Node* top;
      if (left->priority < right->priority) {
        left->propagate();
        left->rch = merge(left->rch, right);
        top = left;
      } else {
        right->propagate();
        right->lch = merge(left, right->lch);
        top = right;
      }
      return top;
    }
    void incr_abv(int key) {
      auto ret = split(key, root);
      if (ret.second) {
        ret.second->lazy++;
      }
      auto cur = merge(ret.first, ret.second);
      root = cur;
    }
    void insert(int key) {
      auto cur = new Node(key);
      auto ret = split(key, root);
      cur = merge(ret.first, cur);
      cur = merge(cur, ret.second);
      root = cur;
    }
    void erase(int key) {
      Node *left, *mid, *right;
      tie(left, mid) = split(key, root);
      tie(mid, right) = split(key + 1, mid);
      root = merge(left, right);
    }
    int size(Node* cur) {
      if (cur == 0) {
        return 0;
      }
      return 1 + size(cur->lch) + size(cur->rch);
    }
    int size() { return size(root); }
    void print(Node* cur) {
      if (cur == 0) {
        return;
      }
      cur->propagate();
      print(cur->lch);
      cerr << cur->value << " ";
      print(cur->rch);
    }
    void print() {
      cerr << " treap : ";
      print(root);
      cerr << endl;
    }
  };
  const int MAX_N = 4e5 + 5;
  Modint fact[MAX_N];
  Modint ifact[MAX_N];
  Modint choose(int n, int k) { return fact[n] * ifact[k] * ifact[n - k]; }
  void solve() {
    int n, m;
    cin >> n >> m;
    Treap breaks;
    for (int i = 0; i < m; i++) {
      int x, y;
      cin >> x >> y;
      breaks.erase(y - 1);
      breaks.incr_abv(y);
      breaks.insert(y);
    }
    cout << choose(2 * n - breaks.size() - 1, n) << '\n';
  }
  int main() {
    fact[0] = 1;
    for (int i = 1; i < MAX_N; i++) {
      fact[i] = fact[i - 1] * i;
    }
    for (int i = 0; i < MAX_N; i++) {
      ifact[i] = inv(fact[i]);
    }
    ios::sync_with_stdio(false);
    cin.tie(0);
    randgen = mt19937(chrono::steady_clock::now().time_since_epoch().count());
    int testc;
    cin >> testc;
    for (int t = 0; t < testc; t++) {
      solve();
    }
  }
  