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 long long INF = 2147483647;
  const long long LNF = INF * INF;
  const long long MOD = 1000000007;
  const long long mod = 998244353;
  const long double eps = 1e-12;
  const long long MAX = 200010;
  long long n, m;
  pair<long long, long long> op[MAX];
  long long fac[2 * MAX];
  long long fpow(long long x, long long p) {
    if (p == 0) return 1;
    if (p % 2) return (x * fpow(x, p - 1)) % mod;
    long long tmp = fpow(x, p / 2);
    return (tmp * tmp) % mod;
  }
  long long inv(long long x) { return fpow(x, mod - 2); }
  struct Treap {
    Treap *l, *r;
    long long pri;
    pair<long long, long long> seg;
    long long tag;
    Treap(long long _L, long long _R)
        : l(nullptr), r(nullptr), pri(rand()), seg(make_pair(_L, _R)), tag(0) {}
  };
  Treap *treap = nullptr;
  long long treap_sz;
  void push(Treap *&T) {
    if (T == nullptr) return;
    T->seg.first += T->tag;
    T->seg.second += T->tag;
    if (T->l != nullptr) T->l->tag += T->tag;
    if (T->r != nullptr) T->r->tag += T->tag;
    T->tag = 0;
  }
  Treap *Merge(Treap *a, Treap *b) {
    if (a == nullptr) return b;
    if (b == nullptr) return a;
    if (a->pri < b->pri) {
      push(a);
      a->r = Merge(a->r, b);
      return a;
    } else {
      push(b);
      b->l = Merge(a, b->l);
      return b;
    }
  }
  void Split_by_L(Treap *T, Treap *&a, Treap *&b, long long k) {
    if (T == nullptr) {
      a = b = nullptr;
      return;
    }
    push(T);
    if (T->seg.first <= k) {
      a = T;
      Split_by_L(a->r, a->r, b, k);
    } else {
      b = T;
      Split_by_L(b->l, a, b->l, k);
    }
  }
  Treap *&Last(Treap *&T) {
    if (T == nullptr) return T;
    push(T);
    if (T->r == nullptr) return T;
    return Last(T->r);
  }
  void Insertion_sort(Treap *&T, long long pos) {
    if (pos == 1) {
      T->tag += 1;
      T = Merge(new Treap(1, 1), T);
      treap_sz += 1;
      return;
    }
    Treap *a, *b, *c;
    Split_by_L(T, a, b, pos - 1);
    Treap *&Node = Last(a);
    if (Node->seg.second > pos - 1) {
      c = new Treap(pos + 1, Node->seg.second + 1);
      treap_sz += 1;
      Node->seg.second = pos;
    } else {
      c = nullptr;
      Node->seg.second = pos;
    }
    if (b != nullptr) b->tag += 1;
    T = Merge(Merge(a, c), b);
  }
  void debug(Treap *&T) {
    if (T == nullptr) return;
    debug(T->l);
    cerr << "[" << T->seg.first << ", " << T->seg.second << "] ";
    debug(T->r);
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    fac[0] = 1;
    for (long long i = (1); i <= (2 * MAX - 1); i += (1)) {
      fac[i] = (fac[i - 1] * i) % mod;
    }
    long long Q;
    cin >> Q;
    while (Q--) {
      cin >> n >> m;
      for (long long i = (1); i <= (m); i += (1)) {
        cin >> op[i].first >> op[i].second;
      }
      treap = new Treap(1, 1);
      treap_sz = 1;
      for (long long i = (1); i <= (m); i += (1)) {
        Treap *&Node = Last(treap);
        Node->seg.second = op[i].first - 1;
        Insertion_sort(treap, op[i].second);
      }
      long long leq = n - treap_sz;
      long long res =
          (((fac[n + leq] * inv(fac[n])) % mod) * inv(fac[leq])) % mod;
      cout << res << '\n';
    }
    return 0;
  }
  