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;
  int mod = 998244353;
  namespace math {
  int inc(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
  int qpow(int a, int x, int mo) {
    int res = 1;
    while (x) {
      if (x & 1) res = 1ll * res * a % mod;
      a = 1ll * a * a % mod;
      x >>= 1;
    }
    return res;
  }
  vector<int> Fac, invFac;
  void Finit(int n) {
    Fac.resize(n + 1);
    invFac.resize(n + 1);
    Fac[0] = 1;
    invFac[0] = 1;
    for (int i = 1; i <= n; i++) Fac[i] = 1ll * Fac[i - 1] * i % mod;
    invFac[n] = qpow(Fac[n], mod - 2, mod);
    for (int i = n - 1; i >= 1; i--)
      invFac[i] = 1ll * invFac[i + 1] * (i + 1) % mod;
  }
  int C(int n, int m) {
    if (m == 0) return 1;
    if (n < m || m < 0) return 0;
    return (int)(1ll * Fac[n] * invFac[m] % mod * invFac[n - m] % mod);
  }
  int A(int n, int m) {
    if (n < m || m < 0) return 0;
    return (int)(1ll * Fac[n] * invFac[n - m] % mod);
  }
  }  // namespace math
  using namespace math;
  std::mt19937 rnd(std::chrono::steady_clock::now().time_since_epoch().count());
  const int N = 1e6 + 10;
  struct FHQ {
    vector<int> ls{0}, rs{0}, siz{0}, key{0}, val{0};
    int tcnt;
    int root;
    int newNode(int x) {
      siz.push_back(1);
      val.push_back(x);
      key.push_back(rnd());
      ls.push_back(0);
      rs.push_back(0);
      return val.size() - 1;
    }
    void pushup(int pos) { siz[pos] = siz[ls[pos]] + siz[rs[pos]] + 1; }
    void split(int pos, int v, int &ll, int &rr) {
      if (!pos) {
        ll = 0, rr = 0;
        return;
      }
      if (siz[ls[pos]] + 1 <= v) {
        ll = pos;
        split(rs[pos], v - siz[ls[pos]] - 1, rs[pos], rr);
      } else {
        rr = pos;
        split(ls[pos], v, ll, ls[pos]);
      }
      pushup(pos);
    }
    int merge(int l, int r) {
      if (!l || !r) return l + r;
      if (key[l] < key[r]) {
        rs[l] = merge(rs[l], r);
        pushup(l);
        return l;
      } else {
        ls[r] = merge(l, ls[r]);
        pushup(r);
        return r;
      }
    }
    void init(int n) {
      for (int i = 1; i <= n; i++) {
        int pos = newNode(i);
        root = merge(root, pos);
      }
    }
  } t;
  void solve() {
    int n, m;
    cin >> n >> m;
    set<int> ans;
    for (int i = 0; i < m; i++) {
      int u, v;
      cin >> u >> v;
      int a, b, c, d, e;
      t.split(t.root, v, b, c);
      t.split(b, v - 1, a, b);
      t.split(c, u - v, c, e);
      t.split(c, u - v - 1, c, d);
      ans.insert(t.val[b]);
      t.root = t.merge(t.merge(a, d), t.merge(b, t.merge(c, e)));
    }
    cout << C(2 * n - 1 - ans.size(), n) << endl;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int M = 2e5;
    math::Finit(M * 2);
    t.init(M);
    int T = 1;
    cin >> T;
    while (T--) {
      solve();
    }
  }
  