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 MAXN = 5e5 + 50, MOD = 998244353, MAXX = 4e5;
  struct SegTree {
    int sum[MAXN << 2];
    int ls(int x) { return x << 1; }
    int rs(int x) { return x << 1 | 1; }
    void push_up(int now) { sum[now] = sum[ls(now)] + sum[rs(now)]; }
    void build(int l, int r, int now) {
      if (l == r) {
        sum[now] = 1;
        return;
      }
      int mid = (l + r) >> 1;
      build(l, mid, ls(now));
      build(mid + 1, r, rs(now));
      push_up(now);
    }
    void change(int x, int k, int l, int r, int now) {
      if (x == l && l == r) {
        sum[now] += k;
        return;
      }
      int mid = (l + r) >> 1;
      if (x <= mid)
        change(x, k, l, mid, ls(now));
      else
        change(x, k, mid + 1, r, rs(now));
      push_up(now);
    }
    int search(int x, int l, int r, int now) {
      if (l == r) return l;
      int mid = (l + r) >> 1;
      if (x <= sum[ls(now)]) return search(x, l, mid, ls(now));
      return search(x - sum[ls(now)], mid + 1, r, rs(now));
    }
  };
  int qpow(int x, int y) {
    int res = 1;
    for (; y; x = 1ll * x * x % MOD, y >>= 1)
      if (y & 1) res = 1ll * res * x % MOD;
    return res;
  }
  int inv(int x) { return qpow(x, MOD - 2); }
  SegTree A;
  queue<int> q;
  int N, M, mul[MAXN], invv[MAXN], Ans = 0;
  pair<int, int> a[MAXN];
  int C(int n, int m) { return 1ll * mul[n] * invv[m] % MOD * invv[n - m] % MOD; }
  void prepare() {
    A.build(1, MAXX, 1);
    mul[0] = invv[0] = 1;
    for (int i = 1; i <= MAXX; ++i) mul[i] = 1ll * mul[i - 1] * i % MOD;
    invv[MAXX] = inv(mul[MAXX]);
    for (int i = MAXX - 1; i >= 1; --i)
      invv[i] = 1ll * invv[i + 1] * (i + 1) % MOD;
  }
  void Init() {
    while (!q.empty()) q.pop();
    scanf("%d%d", &N, &M);
    for (int i = 1; i <= M; ++i) scanf("%d%d", &a[i].first, &a[i].second);
  }
  void Solve() {
    int cnt = 0;
    for (int i = M; i >= 1; --i) {
      int x = A.search(a[i].second, 1, MAXX, 1),
          y = A.search(a[i].second + 1, 1, MAXX, 1);
      if (y == x + 1) ++cnt;
      A.change(x, -1, 1, MAXX, 1);
      q.push(x);
    }
    Ans = C(2 * N - cnt - 1, N);
    while (!q.empty()) {
      A.change(q.front(), 1, 1, MAXX, 1);
      q.pop();
    }
  }
  void Print() { printf("%d\n", Ans); }
  int main() {
    prepare();
    int T;
    cin >> T;
    while (T--) {
      Init();
      Solve();
      Print();
    }
    return 0;
  }
  