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;
  using ll = long long;
  using ld = long double;
  const int mod = 998244353;
  const int inf = 2e9 + 5;
  const ll linf = 9e18 + 5;
  int n, m;
  const int N = 2e5 + 5;
  int arr[N][2];
  ll fact[N * 2];
  ll tcaf[N * 2];
  int seg[N * 3];
  void build(int l = 1, int r = N - 1, int ind = 1) {
    if (l == r) {
      seg[ind] = 1;
      return;
    }
    int m = (l + r) / 2;
    build(l, m, 2 * ind);
    build(m + 1, r, 2 * ind + 1);
    seg[ind] = seg[2 * ind] + seg[2 * ind + 1];
  }
  int findind(int i, int l = 1, int r = N - 1, int ind = 1) {
    if (l == r) {
      return l;
    }
    int m = (l + r) / 2;
    if (seg[2 * ind] >= i) {
      return findind(i, l, m, 2 * ind);
    } else {
      return findind(i - seg[2 * ind], m + 1, r, 2 * ind + 1);
    }
  }
  void update(int i, int x, int l = 1, int r = N - 1, int ind = 1) {
    if (i < l || r < i) {
      return;
    }
    if (l == r) {
      seg[ind] += x;
      return;
    }
    int m = (l + r) / 2;
    update(i, x, l, m, 2 * ind);
    update(i, x, m + 1, r, 2 * ind + 1);
    seg[ind] += x;
  }
  ll mull(ll a, ll b) { return (a * b) % mod; }
  ll binpow(ll a, ll b) {
    if (b == 0) {
      return 1;
    }
    if (b % 2) {
      return mull(a, binpow(a, b - 1));
    }
    ll half = binpow(a, b / 2);
    return mull(half, half);
  }
  ll ncr(ll n, ll r) {
    if (n < r || r < 0) {
      return 0;
    }
    ll d = mull(tcaf[n - r], tcaf[r]);
    return mull(fact[n], d);
  }
  void init() {
    fact[0] = 1;
    for (int i = 1; i < 2 * N; i++) {
      fact[i] = mull(i, fact[i - 1]);
    }
    tcaf[2 * N - 1] = binpow(fact[2 * N - 1], mod - 2);
    for (int i = 2 * N - 2; i > -1; i--) {
      tcaf[i] = mull(i + 1, tcaf[i + 1]);
    }
  }
  void input() {
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
      cin >> arr[i][0] >> arr[i][1];
    }
  }
  void solve() {
    ll top = 2 * n - 1;
    vector<int> nums;
    for (int i = m - 1; i > -1; i--) {
      int x = findind(arr[i][1]);
      int y = findind(arr[i][1] + 1);
      if (x + 1 == y) {
        top--;
      }
      update(x, -1);
      nums.push_back(x);
    }
    for (int i = m - 1; i > -1; i--) {
      update(nums[i], 1);
    }
    cout << ncr(top, n) << endl;
  }
  void output() {}
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    init();
    build();
    int number_of_testcases = 1;
    cin >> number_of_testcases;
    while (number_of_testcases--) {
      input();
      solve();
      output();
    }
  }
  