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;
  inline int read() {
    int x = 0;
    char ch = getchar();
    bool f = 0;
    for (; ch < '0' || ch > '9'; ch = getchar())
      if (ch == '-') f = 1;
    for (; ch >= '0' && ch <= '9'; ch = getchar())
      x = (x << 3) + (x << 1) + (ch ^ 48);
    return f ? -x : x;
  }
  const int N = 200010;
  namespace Treap {
  int son[N][2], val[N], siz[N], rnd[N], cnt, root, tag[N];
  inline void init() { cnt = root = 0; }
  inline int newnode(int v) {
    int u = ++cnt;
    val[u] = v;
    siz[u] = 1;
    rnd[u] = rand() + (rand() << 15);
    son[u][0] = son[u][1] = 0;
    tag[u] = 0;
    return u;
  }
  inline void pushup(int u) { siz[u] = siz[son[u][0]] + siz[son[u][1]] + 1; }
  inline void pushtag(int u, int x) {
    if (!u) return;
    tag[u] += x;
    val[u] += x;
  }
  inline void pushdown(int u) {
    if (!tag[u]) return;
    pushtag(son[u][0], tag[u]);
    pushtag(son[u][1], tag[u]);
    tag[u] = 0;
  }
  int merge(int x, int y) {
    if (!x || !y) return x + y;
    pushdown(x);
    pushdown(y);
    if (rnd[x] < rnd[y]) {
      son[x][1] = merge(son[x][1], y);
      pushup(x);
      return x;
    } else {
      son[y][0] = merge(x, son[y][0]);
      pushup(y);
      return y;
    }
  }
  void split_val(int u, int k, int &x, int &y) {
    if (!u) return void(x = y = 0);
    pushdown(u);
    if (val[u] <= k)
      x = u, split_val(son[x][1], k, son[x][1], y);
    else
      y = u, split_val(son[y][0], k, x, son[y][0]);
    pushup(u);
  }
  void split_rank(int u, int k, int &x, int &y) {
    if (!u) return void(x = y = 0);
    pushdown(u);
    if (k <= siz[son[u][0]])
      y = u, split_rank(son[y][0], k, x, son[y][0]);
    else
      x = u, split_rank(son[x][1], k - siz[son[u][0]] - 1, son[x][1], y);
    pushup(u);
  }
  inline int kth(int u, int k) {
    int s = siz[son[u][0]] + 1;
    if (s == k) return u;
    if (s < k) return kth(son[u][1], k - s);
    return kth(son[u][0], k);
  }
  int x, y, z, ans;
  inline void ins(int v) {
    split_val(root, v - 1, x, y);
    split_val(y, v, y, z);
    if (y) {
      y = merge(y, z);
      pushtag(y, 1);
      root = merge(x, y);
    } else {
      pushtag(z, 1);
      y = merge(newnode(v + 1), z);
      root = merge(x, y);
    }
  }
  inline void del(int v) {
    split_val(root, v, x, y);
    split_val(x, v - 1, x, z);
    root = merge(merge(x, z = merge(son[z][0], son[z][1])), y);
  }
  inline int rank(int v) {
    split_val(root, v - 1, x, y);
    ans = siz[x] + 1;
    root = merge(x, y);
    return ans;
  }
  inline int Kth(int v) { return val[kth(root, v)]; }
  inline int pre(int v) {
    split_val(root, v - 1, x, y);
    ans = val[kth(x, siz[x])];
    root = merge(x, y);
    return ans;
  }
  inline int nex(int v) {
    split_val(root, v, x, y);
    ans = val[kth(y, 1)];
    root = merge(x, y);
    return ans;
  }
  }  // namespace Treap
  const long long mod = 998244353;
  inline long long Add(long long x, long long y) {
    x += y;
    return (x < mod) ? x : x - mod;
  }
  inline long long Dec(long long x, long long y) {
    x -= y;
    return (x < 0) ? x + mod : x;
  }
  inline long long Mul(long long x, long long y) { return x * y % mod; }
  inline long long Pow(long long x, long long y) {
    y %= (mod - 1);
    long long ans = 1;
    for (; y; y >>= 1, x = x * x % mod)
      if (y & 1) ans = ans * x % mod;
    return ans;
  }
  long long fac[N << 1], inv[N << 1];
  void init(int n) {
    fac[0] = 1;
    for (int i = (1), end_i = (n); i <= end_i; i++) fac[i] = fac[i - 1] * i % mod;
    inv[n] = Pow(fac[n], mod - 2);
    for (int i = (n), end_i = (1); i >= end_i; i--) inv[i - 1] = inv[i] * i % mod;
  }
  inline long long C(int n, int m) {
    if (n < 0 || m < 0 || n < m) return 0;
    return fac[n] * inv[m] % mod * inv[n - m] % mod;
  }
  int main() {
    init(4e5);
    int n, m, x, y;
    for (int ___ = (1), end_i = (read()); ___ <= end_i; ___++) {
      Treap::init();
      n = read();
      m = read();
      for (int i = (1), end_i = (m); i <= end_i; i++) {
        x = read();
        y = read();
        Treap::ins(y);
      }
      int k = Treap::cnt;
      printf("%lld\n", C(n + n - k - 1, n));
    }
    return 0;
  }
  