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 N = 4e5 + 1;
  const long long mod = 998244353;
  long long n, m;
  pair<int, int> a[N];
  vector<int> ans;
  int bit[N];
  int rnk[N];
  int s[N];
  inline void upd(int id, int v) {
    for (int i = id; i <= n; i += i & -i) bit[i] += v;
  }
  inline int qry(int id) {
    int res = 0;
    for (int i = id; i >= 1; i -= i & -i) res += bit[i];
    return res;
  }
  inline int get(int x) {
    int cur = 0;
    for (int i = 17; i >= 0; i--) {
      if ((cur + (1 << i)) <= n && x > (1 << i) - bit[cur + (1 << i)]) {
        cur += 1 << i;
        x -= (1 << i) - bit[cur];
      }
    }
    return cur + 1;
  }
  long long f[N], inf[N];
  long long pw(long long x, long long y) {
    if (y == 0) return 1;
    if (y % 2) return x * pw(x, y - 1) % mod;
    long long res = pw(x, y / 2);
    return res * res % mod;
  }
  long long C(long long x, long long y) {
    return f[x] * inf[y] % mod * inf[x - y] % mod;
  }
  long long u[N], v[N];
  void solve() {
    cin >> n >> m;
    for (int i = 1; i <= m; i++) {
      cin >> u[i] >> v[i];
    }
    for (int i = m; i >= 1; i--) {
      a[i].first = get(v[i]);
      a[i].second = i;
      upd(a[i].first, 1);
    }
    for (int i = m; i >= 1; i--) {
      upd(a[i].first, -1);
    }
    sort(a + 1, a + m + 1);
    int cnt = 0;
    for (int i = 1; i <= m; i++) {
      if (i != m && a[i].first + 1 == a[i + 1].first &&
          a[i].second < a[i + 1].second)
        continue;
      else if (a[i].first == n)
        continue;
      else
        cnt++;
    }
    cnt = n - 1 - cnt;
    long long ans = C(cnt + n, cnt);
    cout << ans << '\n';
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    f[0] = 1;
    int iu = 400000;
    for (int i = 1; i <= iu; i++) f[i] = f[i - 1] * i % mod;
    inf[iu] = pw(f[iu], mod - 2);
    for (int i = iu; i >= 1; i--) inf[i - 1] = inf[i] * i % mod;
    int t;
    cin >> t;
    while (t--) solve();
  }
  