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 long long P = 998244353;
  long long T, n, m, fac[400005], inv[400005];
  long long cnt, ans, tu[400005], tv[400005], pos[400005], val[400005];
  long long tr[400005];
  long long lowbit(long long x) { return x & (-x); }
  void modify(long long x, long long v) {
    for (; x <= 400000; x += lowbit(x)) tr[x] += v;
  }
  long long query(long long x) {
    long long ret = 0;
    for (; x; x -= lowbit(x)) ret += tr[x];
    return ret;
  }
  long long pwr(long long x, long long y) {
    long long ret = 1, tmp = x;
    while (y) {
      if (y & 1) ret = (ret * tmp) % P;
      tmp = (tmp * tmp) % P;
      y >>= 1;
    }
    return ret;
  }
  long long comb(long long x, long long y) {
    if (x < y) return 0;
    return fac[x] * inv[y] % P * inv[x - y] % P;
  }
  int main() {
    ios::sync_with_stdio(false);
    fac[0] = 1;
    for (long long i = 1; i <= 400001; i++) fac[i] = (fac[i - 1] * i) % P;
    for (long long i = 0; i <= 400001; i++) inv[i] = pwr(fac[i], P - 2);
    for (long long i = 1; i <= 400000; i++) modify(i, 1);
    cin >> T;
    while (T--) {
      cin >> n >> m;
      ans = cnt = 0;
      for (long long i = 1; i <= m; i++) cin >> tv[i] >> tu[i];
      for (long long l, r, mid, i = m; i >= 1; i--) {
        l = 1;
        r = n;
        while (l <= r) {
          mid = (l + r) >> 1;
          if (query(mid) >= tu[i]) {
            pos[i] = mid;
            r = mid - 1;
          } else
            l = mid + 1;
        }
        modify(pos[i], -1);
        val[pos[i]] = tv[i];
      }
      for (long long i = 1; i <= m; i++) {
        if (pos[i] != n) {
          if (!val[pos[i] + 1])
            cnt++;
          else if (val[pos[i] + 1] < val[pos[i]])
            cnt++;
        }
      }
      for (long long i = m; i >= 1; i--) {
        val[pos[i]] = 0;
        modify(pos[i], 1);
        pos[i] = 0;
      }
      ans = comb(n * 2 - 1 - cnt, n);
      cout << ans << '\n';
    }
    return 0;
  }
  