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;
  using ll = long long;
  const int mod = 998244353;
  int qpow(int a, int n) {
    int ans = 1;
    for (; n; n >>= 1, a = (ll)a * a % mod)
      if (n & 1) ans = (ll)ans * a % mod;
    return ans;
  }
  const int maxn = 400400;
  int fac[maxn], ifac[maxn];
  int C(int n, int m) {
    return fac[n] * (ll)ifac[m] % mod * (ll)ifac[n - m] % mod;
  }
  struct node {
    int val, tag;
    int pa, ch[2];
  } buf[maxn];
  int tot, rt;
  int init(int v) {
    buf[++tot] = (node){v, 0, 0, 0, 0};
    return tot;
  }
  void push_down(int rt) {
    if (buf[rt].tag) {
      if (buf[rt].ch[0]) {
        buf[buf[rt].ch[0]].val += buf[rt].tag;
        buf[buf[rt].ch[0]].tag += buf[rt].tag;
      }
      if (buf[rt].ch[1]) {
        buf[buf[rt].ch[1]].val += buf[rt].tag;
        buf[buf[rt].ch[1]].tag += buf[rt].tag;
      }
      buf[rt].tag = 0;
    }
  }
  void rot(int x, int c) {
    int g = buf[buf[x].pa].pa;
    int p = buf[x].pa;
    int b = buf[x].ch[!c];
    push_down(g);
    push_down(p);
    push_down(x);
    if (g) {
      if (buf[g].ch[0] == p)
        buf[g].ch[0] = x;
      else
        buf[g].ch[1] = x;
    }
    buf[x].pa = g;
    buf[x].ch[!c] = p;
    buf[p].pa = x;
    buf[p].ch[c] = b;
    if (b) buf[b].pa = p;
  }
  void splay(int x) {
    while (buf[x].pa) {
      int p = buf[x].pa, g = buf[p].pa;
      if (g) {
        if (x == buf[p].ch[0] && p == buf[g].ch[0]) {
          rot(p, 0);
          rot(x, 0);
        } else if (x == buf[p].ch[0] && p == buf[g].ch[1]) {
          rot(x, 0);
          rot(x, 1);
        } else if (x == buf[p].ch[1] && p == buf[g].ch[0]) {
          rot(x, 1);
          rot(x, 0);
        } else {
          rot(p, 1);
          rot(x, 1);
        }
      } else {
        if (x == buf[p].ch[0])
          rot(x, 0);
        else
          rot(x, 1);
      }
    }
  }
  int ask(int v) {
    int ans = 0;
    int u = rt;
    while (u) {
      push_down(u);
      if (buf[u].val >= v) {
        ans = u;
        u = buf[u].ch[0];
      } else {
        u = buf[u].ch[1];
      }
    }
    if (ans) {
      splay(ans);
      rt = ans;
    }
    return ans;
  }
  int main(void) {
    int N = 400004;
    fac[0] = 1;
    for (int i = 1; i <= N; i++) fac[i] = fac[i - 1] * (ll)i % mod;
    ifac[N] = qpow(fac[N], mod - 2);
    for (int i = N - 1; i >= 0; i--) ifac[i] = ifac[i + 1] * (ll)(i + 1) % mod;
    int T;
    scanf("%d", &T);
    while (T--) {
      int n, m;
      scanf("%d%d", &n, &m);
      tot = rt = 0;
      int cnt = 0;
      for (int i = 0; i < m; i++) {
        int x, y;
        scanf("%d%d", &x, &y);
        int u = ask(y);
        if (buf[u].val == y) {
          buf[u].val++;
          if (buf[u].ch[1]) {
            buf[buf[u].ch[1]].val++;
            buf[buf[u].ch[1]].tag++;
          }
          continue;
        }
        cnt++;
        int v = init(y);
        if (!u) {
          buf[v].ch[0] = rt;
          if (rt) buf[rt].pa = v;
          rt = v;
        } else {
          push_down(u);
          int lch = buf[u].ch[0];
          buf[u].ch[0] = 0;
          buf[v].ch[0] = lch;
          buf[v].ch[1] = u;
          if (lch) buf[lch].pa = v;
          if (u) buf[u].pa = v;
          rt = v;
        }
        buf[v].val++;
        if (buf[v].ch[1]) {
          buf[buf[v].ch[1]].val++;
          buf[buf[v].ch[1]].tag++;
        }
      }
      printf("%d\n", C(n + n - 1 - cnt, n));
    }
    return 0;
  }
  