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;
  void read(int &x) {
    x = 0;
    int f = 1;
    char ch = getchar();
    for (; !isdigit(ch); ch = getchar())
      if (ch == '-') f = -f;
    for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
    x *= f;
  }
  void print(int x) {
    if (x < 0) putchar('-'), x = -x;
    if (!x) return;
    print(x / 10), putchar(x % 10 + 48);
  }
  void write(int x) {
    if (!x)
      putchar('0');
    else
      print(x);
    putchar('\n');
  }
  const int maxn = 4e5 + 10;
  const int inf = 1e9;
  const double eps = 1e-8;
  const int mod = 998244353;
  int n, m, a[maxn], t[maxn << 2], N = 4e5;
  void build(int p, int l, int r) {
    if (l == r) return t[p] = 1, void();
    build(p << 1, l, ((l + r) >> 1)), build(p << 1 | 1, ((l + r) >> 1) + 1, r);
    t[p] = t[p << 1] + t[p << 1 | 1];
  }
  void fuck(int p, int l, int r, int x) {
    if (l == r) return t[p] ^= 1, void();
    if (x <= ((l + r) >> 1))
      fuck(p << 1, l, ((l + r) >> 1), x);
    else
      fuck(p << 1 | 1, ((l + r) >> 1) + 1, r, x);
    t[p] = t[p << 1] + t[p << 1 | 1];
  }
  int find(int p, int l, int r, int k) {
    if (l == r) return l;
    if (t[p << 1] >= k)
      return find(p << 1, l, ((l + r) >> 1), k);
    else
      return find(p << 1 | 1, ((l + r) >> 1) + 1, r, k - t[p << 1]);
  }
  int sta[maxn], top, fac[maxn], ifac[maxn];
  int qpow(int a, int x) {
    int res = 1;
    for (; x; x >>= 1, a = 1ll * a * a % mod)
      if (x & 1) res = 1ll * res * a % mod;
    return res;
  }
  void prepare() {
    fac[0] = ifac[0] = 1;
    for (int i = 1; i <= N; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
    ifac[N] = qpow(fac[N], mod - 2);
    for (int i = N - 1; i; i--) ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
  }
  int c(int a, int b) { return 1ll * fac[a] * ifac[b] % mod * ifac[a - b] % mod; }
  void solve() {
    read(n), read(m);
    int cnt = 0, ans = 0;
    for (int i = 1; i <= m; i++) read(a[i]), read(a[i]);
    for (int i = m; i; i--) {
      int x = find(1, 1, N, a[i]), nxt = find(1, 1, N, a[i] + 1);
      fuck(1, 1, N, x), sta[++top] = x;
      if (nxt == x + 1) cnt++;
    }
    while (top) fuck(1, 1, N, sta[top--]);
    ans = c(2 * n - cnt - 1, n - cnt - 1);
    write(ans);
  }
  int main() {
    build(1, 1, N);
    prepare();
    int t;
    read(t);
    while (t--) solve();
    return 0;
  }
  