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 mod = 998244353;
  inline void add(int &x, const int y) { (x += y) >= mod && (x -= mod); }
  inline int Add(int x, const int y) { return add(x, y), x; }
  inline void sub(int &x, const int y) { (x -= y) < 0 && (x += mod); }
  inline int Sub(int x, const int y) { return sub(x, y), x; }
  inline void mul(int &x, const int y) { x = 1ll * x * y % mod; }
  inline int Mul(const int x, const int y) { return 1ll * x * y % mod; }
  inline int power(int a, int b, int c = 1) {
    for (; b; b >>= 1, mul(a, a))
      if (b & 1) mul(c, a);
    return c;
  }
  const int N = 555555;
  mt19937 rng(std::chrono::steady_clock::now().time_since_epoch().count());
  int inv[N], fac[N], ifac[N];
  int ch[N][2], siz[N], lazy[N], val[N], pri[N], root, tot;
  inline void tag(int p, int val) {
    ::val[p] += val;
    lazy[p] += val;
  }
  inline void pushdown(int p) {
    if (lazy[p]) {
      if (ch[p][0]) tag(ch[p][0], lazy[p]);
      if (ch[p][1]) tag(ch[p][1], lazy[p]);
      lazy[p] = 0;
    }
  }
  inline void pushup(int p) { siz[p] = 1 + siz[ch[p][0]] + siz[ch[p][1]]; }
  void split(int p, int v, int &x, int &y) {
    if (!p)
      x = y = 0;
    else {
      pushdown(p);
      if (val[p] <= v)
        x = p, split(ch[p][1], v, ch[p][1], y);
      else
        y = p, split(ch[p][0], v, x, ch[p][0]);
      pushup(p);
    }
  }
  int merge(int x, int y) {
    if (!x || !y) return x | y;
    pushdown(x), pushdown(y);
    if (pri[x] > pri[y]) {
      ch[x][1] = merge(ch[x][1], y);
      pushup(x);
      return x;
    } else {
      ch[y][0] = merge(x, ch[y][0]);
      pushup(y);
      return y;
    }
  }
  inline void insert(int v) {
    int x, y;
    split(root, v, x, y);
    siz[++tot] = 1, val[tot] = v;
    pri[tot] = rng();
    ch[tot][0] = ch[tot][1] = lazy[tot] = 0;
    root = merge(merge(x, tot), y);
  }
  inline void erase(int v) {
    int x, y;
    split(root, v, x, y);
    int a, b;
    split(x, v - 1, a, b);
    b = merge(ch[b][0], ch[b][1]);
    root = merge(merge(a, b), y);
  }
  inline int rnk(int v) {
    int x, y;
    split(root, v - 1, x, y);
    int res = siz[x] + 1;
    root = merge(x, y);
    return res;
  }
  inline void add(int pos) {
    int x, y;
    split(root, pos - 1, x, y);
    tag(y, 1);
    root = merge(x, y);
  }
  inline void init(int n) {
    for (int i = 0; i < 2; ++i) inv[i] = fac[i] = ifac[i] = 1;
    for (int i = 2; i <= n; ++i) {
      inv[i] = Mul(mod - mod / i, inv[mod % i]);
      fac[i] = Mul(fac[i - 1], i);
      ifac[i] = Mul(ifac[i - 1], inv[i]);
    }
  }
  inline int binom(int n, int m) {
    if (n < 0 || m < 0 || n < m) return 0;
    return Mul(fac[n], Mul(ifac[m], ifac[n - m]));
  }
  inline int calc(int n, int m, int k) {
    k = n + 1 - k;
    return binom(m - 2 + k, n);
  }
  void solve() {
    int n, m;
    scanf("%d%d", &n, &m);
    tot = root = 0;
    for (int i = 1; i <= m; ++i) {
      int x, y;
      scanf("%d%d", &x, &y);
      add(y);
      erase(y - 1);
      erase(y);
      insert(y);
    }
    printf("%d\n", calc(n, n, siz[root]));
  }
  int main() {
    init(N - 1);
    int t;
    scanf("%d", &t);
    while (t--) solve();
    return 0;
  }
  