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 MX = 200200, BS = 522, md = 998244353;
  int t, n, m, num, cnt, x, y, i, j, le, ri, mid, prv, stptr, ptr, bptr, nptr,
      nxt[BS], add[BS], st[BS], fin[BS], a[MX], b[MX];
  long long f[2 * MX], o[2 * MX];
  long long pw(long long a, int b) {
    if (b == 0) return 1LL;
    if (b & 1) return (pw(a, b - 1) * a) % md;
    long long x = pw(a, b / 2);
    return (x * x) % md;
  }
  void createbptr() {
    bptr = num++;
    add[bptr] = 0;
    st[bptr] = fin[bptr] = cnt;
    a[cnt++] = y;
  }
  int main() {
    for (f[0] = o[0] = i = 1; i < 2 * MX; ++i) {
      f[i] = (f[i - 1] * i) % md;
      o[i] = pw(f[i], md - 2);
    }
    scanf("%d", &t);
    while (t--) {
      scanf("%d%d", &n, &m);
      if (m == 0) {
        printf("%d\n", int((((f[n + n - 1] * o[n]) % md) * o[n - 1]) % md));
        continue;
      }
      scanf("%d%d", &x, &y);
      stptr = 0;
      a[0] = ++y;
      add[0] = st[0] = fin[0] = 0;
      nxt[0] = -1;
      num = 1;
      cnt = 1;
      for (i = 1; i < m; i++) {
        scanf("%d%d", &x, &y);
        prv = -1;
        for (ptr = stptr; ptr != -1; ptr = nxt[ptr]) {
          x = y - add[ptr];
          if (x <= a[fin[ptr]]) {
            if (x <= a[st[ptr]]) {
              if (x < a[st[ptr]]) {
                createbptr();
                if (prv == -1)
                  stptr = bptr;
                else
                  nxt[prv] = bptr;
                nxt[bptr] = ptr;
                ptr = bptr;
              }
            } else {
              le = st[ptr];
              ri = fin[ptr];
              while (le < ri) {
                mid = (le + ri) / 2;
                if (x <= a[mid])
                  ri = mid;
                else
                  le = mid + 1;
              }
              nptr = num++;
              st[nptr] = ri;
              fin[nptr] = fin[ptr];
              add[nptr] = add[ptr];
              nxt[nptr] = nxt[ptr];
              fin[ptr] = ri - 1;
              if (x < a[ri]) {
                createbptr();
                nxt[ptr] = bptr;
                nxt[bptr] = nptr;
              } else
                nxt[ptr] = nptr;
              ptr = nxt[ptr];
            }
            break;
          }
          prv = ptr;
        }
        if (ptr == -1) {
          createbptr();
          if (prv == -1)
            stptr = bptr;
          else
            nxt[prv] = bptr;
          nxt[bptr] = -1;
          ptr = bptr;
        }
        for (; ptr != -1; ptr = nxt[ptr]) ++add[ptr];
        if (num >= BS - 2) {
          cnt = 0;
          for (ptr = stptr; ptr != -1; ptr = nxt[ptr])
            for (j = st[ptr]; j <= fin[ptr]; ++j) b[cnt++] = a[j] + add[ptr];
          for (j = 0; j < cnt; j++) a[j] = b[j];
          stptr = 0;
          add[0] = st[0] = 0;
          fin[0] = cnt - 1;
          nxt[0] = -1;
          num = 1;
        }
      }
      cnt = n - 1 - cnt;
      printf("%d\n", int((((f[n + cnt] * o[cnt]) % md) * o[n]) % md));
    }
    return 0;
  }
  