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 INFi = 2e9 + 5;
  const int md = 998244353;
  const long long INF = 2e18;
  const int maxN = 3e6 + 1;
  double getTime() { return clock() / (double)CLOCKS_PER_SEC; }
  inline int add(const int &a, const int &b) {
    return a + b >= md ? a + b - md : a + b;
  }
  inline int sub(const int &a, const int &b) {
    return a - b < 0 ? a - b + md : a - b;
  }
  inline int mul(const int &a, const int &b) { return (1ll * a * b) % md; }
  int binpow(int a, int b) {
    if (b <= 0) return 1;
    if (b % 2) return mul(a, binpow(a, b - 1));
    int m = binpow(a, b / 2);
    return mul(m, m);
  }
  int rev(int a) { return binpow(a, md - 2); }
  const int maxF = 4e5 + 5;
  int fact[maxF];
  int rfact[maxF];
  void init() {
    fact[0] = 1;
    for (int i = 1; i < maxF; ++i) fact[i] = mul(fact[i - 1], i);
    rfact[maxF - 1] = rev(fact[maxF - 1]);
    for (int i = maxF - 1; i >= 1; --i) {
      rfact[i - 1] = mul(rfact[i], i);
    }
  }
  int C(int n, int k) {
    if (k < 0 || n < k) return 0;
    return mul(fact[n], mul(rfact[k], rfact[n - k]));
  }
  int T(int n, int k) { return C(n + k - 1, k - 1); }
  struct fenwick {
    int n{};
    vector<long long> fenw{};
    void build(int k) {
      n = k;
      fenw.resize(n);
    }
    void upd(int i, long long x) {
      for (; i < n; i = i | (i + 1)) fenw[i] += x;
    }
    long long get(int i) {
      long long res = 0;
      for (; i >= 0; i = (i & (i + 1)) - 1) res += fenw[i];
      return res;
    }
    long long get(int l, int r) {
      if (l > r) return 0;
      return get(r) - get(l - 1);
    }
    int lower_bound(long long x) {
      int r = -1;
      for (int j = 19; j >= 0; --j) {
        int w = 1 << j;
        if (r + w < n && fenw[r + w] < x) {
          r += w;
          x -= fenw[r];
        }
      }
      return r + 1;
    }
  } F;
  set<int> alive;
  void solve() {
    int n, m;
    cin >> n >> m;
    vector<int> x(m), y(m);
    for (int i = 0; i < (m); ++i) cin >> x[i] >> y[i];
    set<int> bad;
    vector<int> mem1, mem2;
    for (int j = m - 1; j >= 0; --j) {
      int i = F.lower_bound(y[j] - 1);
      int pos = *alive.lower_bound(i);
      int pos2 = *alive.upper_bound(pos);
      auto it = bad.lower_bound(pos);
      if (it == bad.end() || (*it) >= pos2) {
        bad.insert(pos);
      }
      alive.erase(pos);
      F.upd(i, -1);
      mem1.push_back(pos);
      mem2.push_back(i);
    }
    for (auto &w : mem1) alive.insert(w);
    for (auto &w : mem2) F.upd(w, 1);
    cout << T(n - 1 - (int)bad.size(), n + 1) << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int tests = 1;
    init();
    F.build(2e5 + 10);
    for (int i = 0; i < (2e5 + 9); ++i) {
      F.upd(i + 1, 1);
    }
    for (int i = 0; i < (2e5 + 10); ++i) alive.insert(i);
    cin >> tests;
    for (int _ = 0; _ < (tests); ++_) {
      solve();
    }
    return 0;
  }
  