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;
  template <typename T>
  void read(T &x) {
    T f = 1;
    x = 0;
    char s = getchar();
    while (s < '0' || s > '9') {
      if (s == '-') f = -1;
      s = getchar();
    }
    while (s >= '0' && s <= '9') {
      x = (x << 3) + (x << 1) + (s ^ '0');
      s = getchar();
    }
    x *= f;
  }
  template <typename T>
  void write(T x, char s = '\n') {
    if (x < 0) {
      putchar('-');
      x = -x;
    }
    if (!x) {
      putchar('0');
      putchar(s);
      return;
    }
    T tmp[25] = {}, t = 0;
    while (x) tmp[t++] = x % 10, x /= 10;
    while (t-- > 0) putchar(tmp[t] + '0');
    putchar(s);
  }
  const int MAXN = 2e5 + 5;
  const int mod = 998244353;
  const int N = 2e5;
  long long qpow(long long a, long long b) {
    long long res = 1;
    while (b) {
      if (b & 1) res = res * a % mod;
      a = a * a % mod;
      b >>= 1;
    }
    return res;
  }
  int tr[MAXN << 2];
  long long fac[MAXN << 1], finv[MAXN << 1];
  long long C(long long n, long long m) {
    return fac[n] * finv[n - m] % mod * finv[m] % mod;
  }
  void build(int l, int r, int now) {
    if (l == r) {
      tr[now] = 1;
      return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, now << 1);
    build(mid + 1, r, now << 1 | 1);
    tr[now] = (tr[now << 1] + tr[now << 1 | 1]);
  }
  int find(int l, int r, int now, int v) {
    if (l == r) {
      tr[now] = 0;
      return l;
    }
    int mid = (l + r) >> 1;
    if (tr[now << 1] >= v) {
      int res = find(l, mid, now << 1, v);
      tr[now] = (tr[now << 1] + tr[now << 1 | 1]);
      return res;
    } else {
      int res = find(mid + 1, r, now << 1 | 1, v - tr[now << 1]);
      tr[now] = (tr[now << 1] + tr[now << 1 | 1]);
      return res;
    }
  }
  void update(int l, int r, int now) {
    if (l == r) {
      tr[now] = 1;
      return;
    }
    int mid = (l + r) >> 1;
    if (tr[now << 1] != (mid - l + 1)) update(l, mid, now << 1);
    if (tr[now << 1 | 1] != (r - mid)) update(mid + 1, r, now << 1 | 1);
    tr[now] = r - l + 1;
  }
  vector<int> del;
  int vis[MAXN << 1], X[MAXN], Y[MAXN];
  int main() {
    build(1, N, 1);
    fac[0] = 1;
    for (int i = 1; i <= N * 2; ++i) fac[i] = fac[i - 1] * i % mod, vis[i] = 1;
    finv[N * 2] = qpow(fac[N * 2], mod - 2);
    for (int i = N * 2 - 1; i >= 0; --i) finv[i] = finv[i + 1] * (i + 1) % mod;
    int T;
    read(T);
    while (T-- > 0) {
      long long n, m;
      read(n), read(m);
      int ans = n - 1;
      for (int i = 1; i <= m; ++i) read(X[i]), read(Y[i]);
      for (int i = m; i >= 1; --i) {
        long long x = X[i], y = Y[i];
        int t = find(1, N, 1, y);
        vis[t] = 0;
        if (vis[t + 1]) ans--;
        del.push_back(t);
      }
      write(C(ans + n, n));
      for (int i = 0; i < (int)del.size(); ++i) {
        int x = del[i];
        vis[x] = 1;
      }
      del.clear();
      update(1, N, 1);
    }
    return 0;
  }
  