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 = 400005;
  const int MOD = 998244353;
  long long modpow(long long x, long long p) {
    if (p == 0) return 1;
    if (p % 2 == 1) return modpow(x, p - 1) * x % MOD;
    long long r = modpow(x, p / 2);
    return r * r % MOD;
  }
  long long modinv(long long x) { return modpow(x, MOD - 2); }
  long long fact[MAXN], invfact[MAXN];
  long long C(int n, int r) {
    return fact[n] * invfact[r] % MOD * invfact[n - r] % MOD;
  }
  int main() {
    fact[0] = 1;
    for (int i = 1; i < MAXN; i++) fact[i] = i * fact[i - 1] % MOD;
    invfact[MAXN - 1] = modinv(fact[MAXN - 1]);
    for (int i = MAXN - 2; i >= 0; i--)
      invfact[i] = invfact[i + 1] * (i + 1) % MOD;
    int t;
    scanf("%d", &t);
    while (t--) {
      int n, m;
      scanf("%d %d", &n, &m);
      vector<vector<int> > buckets(1);
      vector<int> sum(1, 1);
      buckets[0].push_back(1);
      int last = 1, cnt = 0;
      while (m--) {
        int x, y;
        scanf("%d %d", &x, &y);
        buckets.back().back() += x - last - 1;
        sum.back() += x - last - 1;
        if (y == 1) {
          buckets[0].insert(buckets[0].begin(), 1);
          sum[0]++;
        } else {
          y--;
          for (int i = 0; i < (int)buckets.size(); i++) {
            if (sum[i] < y) {
              y -= sum[i];
            } else {
              sum[i]++;
              for (int j = 0; j < (int)buckets[i].size(); j++) {
                if (buckets[i][j] < y) {
                  y -= buckets[i][j];
                } else {
                  if (buckets[i][j] == y) {
                    buckets[i][j]++;
                  } else {
                    buckets[i].insert(buckets[i].begin() + (j + 1),
                                      buckets[i][j] - y);
                    buckets[i][j] = y + 1;
                  }
                  break;
                }
              }
              break;
            }
          }
        }
        last = x;
        cnt++;
        if (cnt == 300) {
          vector<int> v;
          for (int i = 0; i < (int)buckets.size(); i++)
            for (int j : buckets[i]) v.push_back(j);
          buckets.clear();
          sum.clear();
          buckets.push_back(vector<int>());
          sum.push_back(0);
          for (int j : v) {
            if (buckets.back().size() == 300) {
              buckets.push_back(vector<int>());
              sum.push_back(0);
            }
            buckets.back().push_back(j);
            sum.back() += j;
          }
          cnt = 0;
        }
      }
      buckets.back().back() += n - last;
      vector<int> v;
      for (int i = 0; i < (int)buckets.size(); i++)
        for (int j : buckets[i]) v.push_back(j);
      int rem = n - ((int)v.size());
      printf("%lld\n", C(rem + n, n));
    }
    return 0;
  }
  