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 mod = 998244353;
  const int maxn = 200020;
  struct Splay {
    long long rt, val[maxn], tp[maxn], fa[maxn], ch[maxn][2], tg[maxn], nodecnt;
    void push_down(long long x) {
      val[x] += tg[x];
      if (ch[x][0] != -1) tg[ch[x][0]] += tg[x];
      if (ch[x][1] != -1) tg[ch[x][1]] += tg[x];
      tg[x] = 0;
      return;
    }
    long long son_type(long long x) { return (ch[fa[x]][1] == x); }
    void del_node(long long x) {
      fa[x] = ch[x][0] = ch[x][1] = -1, tp[x] = val[x] = tg[x] = 0;
      return;
    }
    void rotate(long long x) {
      long long y = fa[x], z = fa[y], tp = son_type(x);
      ch[y][tp] = ch[x][tp ^ 1];
      if (ch[x][tp ^ 1] != -1) fa[ch[x][tp ^ 1]] = y;
      ch[x][tp ^ 1] = y;
      fa[y] = x, fa[x] = z;
      if (z != -1) ch[z][ch[z][1] == y] = x;
      return;
    }
    void splay(long long x, long long dest) {
      long long f = fa[x];
      while (f != dest) {
        if (fa[f] == dest) {
          rotate(x);
          break;
        }
        if (son_type(x) == son_type(f))
          rotate(f), rotate(x), f = fa[x];
        else
          rotate(x), rotate(x), f = fa[x];
      }
      if (dest == -1) rt = x;
      return;
    }
    void ins(long long x) {
      long long cur = rt, f;
      if (rt == -1) {
        rt = nodecnt, del_node(nodecnt), fa[nodecnt] = -1, val[nodecnt] = x,
        nodecnt++;
        return;
      }
      while (true) {
        push_down(cur);
        f = cur, cur = ch[cur][val[cur] < x];
        if (cur == -1) {
          del_node(nodecnt), fa[nodecnt] = f, ch[f][val[f] < x] = nodecnt,
                             val[nodecnt] = x, splay(nodecnt, -1), nodecnt++;
          return;
        }
      }
      return;
    }
    long long qry_next() {
      long long cur = ch[rt][1];
      while (cur != -1) {
        if (ch[cur][0] == -1) {
          splay(cur, -1);
          return cur;
        }
        cur = ch[cur][0];
      }
      return 1000000000;
    }
    long long fnd(long long x) {
      long long cur = rt;
      while (cur != -1) {
        push_down(cur);
        if (val[cur] == x) {
          splay(cur, -1);
          return cur;
        }
        cur = ch[cur][val[cur] < x];
      }
      return -1;
    }
    void upd_one(long long x) {
      splay(x, -1);
      val[rt]++;
      if (ch[rt][1] != -1) tg[ch[rt][1]]++;
      return;
    }
  } spl;
  long long n, m, dx[maxn], dy[maxn], fac[maxn << 1], inv[maxn << 1],
      inv2[maxn << 1];
  long long C(long long x, long long y) {
    return (x < 0 || y < 0 || x < y)
               ? 0
               : (fac[x] * ((inv2[y] * inv2[x - y]) % mod)) % mod;
  }
  int main() {
    long long t, tot;
    fac[0] = fac[1] = inv[1] = inv2[0] = inv2[1] = 1;
    for (int i = 2; i < (maxn << 1); i++)
      fac[i] = (fac[i - 1] * i) % mod,
      inv[i] = (mod - (((mod / i) * inv[mod % i]) % mod)) % mod,
      inv2[i] = (inv2[i - 1] * inv[i]) % mod;
    scanf("%lld", &t);
    for (int testcases = 0; testcases < t; testcases++) {
      for (int i = 0; i < spl.nodecnt; i++) spl.del_node(i);
      spl.nodecnt = 0, spl.rt = -1;
      scanf("%lld%lld", &n, &m);
      for (int i = 0; i < m; i++) scanf("%lld%lld", &dx[i], &dy[i]);
      tot = 0;
      for (int i = 0; i < m; i++) {
        long long plc = spl.fnd(dy[i] - 1);
        if (plc == -1 || spl.tp[plc] == 0)
          tot++;
        else
          spl.tp[plc] = 0;
        plc = spl.fnd(dy[i]);
        if (plc == -1) {
          spl.ins(dy[i]);
          plc = spl.fnd(dy[i]);
          spl.tp[plc] = 1;
          plc = spl.qry_next();
          if (plc != 1000000000) spl.upd_one(plc);
        } else {
          spl.upd_one(plc);
          spl.ins(dy[i]);
          plc = spl.fnd(dy[i]);
          spl.tp[plc] = 1;
        }
      }
      printf("%lld\n", C(n + n - tot - 1, n));
    }
    return 0;
  }
  