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;
  long long t, n, m, num, x[200005], y[200005], fac[200005 << 1],
      invfac[200005 << 1];
  set<long long> ans, fucked;
  struct segtree {
    long long l, r, siz;
  } sgt[200005 << 2];
  inline long long read() {
    long long s = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      s = (s << 1) + (s << 3) + ch - '0';
      ch = getchar();
    }
    return s * f;
  }
  void build(long long p, long long l, long long r) {
    sgt[p].l = l, sgt[p].r = r;
    if (l == r) {
      sgt[p].siz = 1;
      return;
    }
    long long mid = l + r >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    sgt[p].siz = sgt[p << 1].siz + sgt[p << 1 | 1].siz;
  }
  void upd(long long p, long long x, long long y) {
    long long l = sgt[p].l, r = sgt[p].r, mid = l + r >> 1;
    if (l == r) {
      sgt[p].siz = y;
      return;
    }
    if (x > mid)
      upd(p << 1 | 1, x, y);
    else
      upd(p << 1, x, y);
    sgt[p].siz = sgt[p << 1].siz + sgt[p << 1 | 1].siz;
  }
  long long find(long long p, long long k) {
    long long l = sgt[p].l, r = sgt[p].r, mid = l + r >> 1;
    if (l == r) return l;
    if (sgt[p << 1].siz >= k) return find(p << 1, k);
    return find(p << 1 | 1, k - sgt[p << 1].siz);
  }
  void exgcd(long long a, long long b, long long &x, long long &y) {
    if (b == 0) {
      x = 1, y = 0;
      return;
    }
    exgcd(b, a % b, y, x);
    y -= (a / b) * x;
  }
  long long inv(long long a, long long p) {
    long long x, y;
    exgcd(a, p, x, y);
    return (x % p + p) % p;
  }
  long long C(long long n, long long m) {
    return fac[n] * invfac[m] % 998244353 * invfac[n - m] % 998244353;
  }
  void init() {
    fac[0] = 1;
    for (long long i = 1; i <= 400000; ++i) fac[i] = fac[i - 1] * i % 998244353;
    invfac[400000] = inv(fac[400000], 998244353);
    for (long long i = 400000; i; --i) invfac[i - 1] = invfac[i] * i % 998244353;
  }
  signed main() {
    init();
    t = read();
    build(1, 1, 200000);
    while (t--) {
      n = read(), m = read(), ans.clear(), fucked.clear();
      for (long long i = 1; i <= m; ++i) x[i] = read(), y[i] = read();
      for (long long i = m, p; i >= 1; --i) {
        p = find(1, y[i]);
        upd(1, p, 0);
        fucked.insert(p);
        p = find(1, y[i]);
        ans.insert(p);
      }
      for (set<long long>::iterator it = fucked.begin(); it != fucked.end(); ++it)
        upd(1, (*it), 1);
      num = n - 1 - ans.size();
      printf("%lld\n", C(n + num, n));
    }
    return 0;
  }
  