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>
  const int N = 200200;
  using namespace std;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const int Q = 400200;
  const int mod = 998244353;
  int pw(int x, int n) {
    int res = 1;
    for (; n > 0; n >>= 1) {
      if (n & 1) {
        res = 1ll * res * x % mod;
      }
      x = 1ll * x * x % mod;
    }
    return res;
  }
  int fac[Q], inv[Q];
  void make() {
    fac[0] = 1;
    for (int i = 1; i < Q; i++) fac[i] = fac[i - 1] * 1ll * i % mod;
    inv[Q - 1] = pw(fac[Q - 1], mod - 2);
    for (int i = Q - 2; i >= 0; i--) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
  }
  int C(int n, int k) {
    if (k < 0 || k > n) return 0;
    return 1ll * fac[n] * inv[k] % mod * inv[n - k] % mod;
  }
  void add(int &x, int y) {
    x += y;
    if (x >= mod) x -= mod;
  }
  void sub(int &x, int y) {
    x -= y;
    if (x < 0) x += mod;
  }
  struct treap {
    treap *l, *r;
    int x, y, delta;
    treap(int _x) {
      l = r = NULL;
      x = _x;
      y = rng();
      delta = 0;
    }
  };
  typedef treap *tree;
  tree root;
  void push(tree v) {
    if (v == NULL || v->delta == 0) {
      return;
    }
    if (v->l != NULL) {
      v->l->x += v->delta;
      v->l->delta += v->delta;
    }
    if (v->r != NULL) {
      v->r->x += v->delta;
      v->r->delta += v->delta;
    }
    v->delta = 0;
  }
  void split(tree v, tree &l, tree &r, int x) {
    if (v == NULL) {
      l = r = NULL;
      return;
    }
    push(v);
    if (v->x <= x) {
      split(v->r, v->r, r, x);
      l = v;
    } else {
      split(v->l, l, v->l, x);
      r = v;
    }
  }
  tree merg(tree l, tree r) {
    if (l == NULL || r == NULL) {
      return l == NULL ? r : l;
    }
    push(l);
    push(r);
    if (l->y > r->y) {
      l->r = merg(l->r, r);
      return l;
    } else {
      r->l = merg(l, r->l);
      return r;
    }
  }
  void prnt(tree v) {
    if (v == NULL) {
      return;
    }
    prnt(v->l);
    cout << v->x << " ";
    prnt(v->r);
  }
  int n, m;
  int X[N];
  int id[N];
  int b[N];
  int t[N];
  void solve() {
    root = NULL;
    cin >> n >> m;
    int res = 2 * n - 1 - m;
    for (int i = 1; i <= m; i++) {
      int x, y;
      cin >> x >> y;
      y -= 1;
      tree a, b, c;
      split(root, a, c, y);
      split(a, a, b, y - 1);
      if (c != NULL) {
        c->delta += 1;
        c->x += 1;
      }
      if (b != NULL) {
        res += 1;
      }
      b = new treap(y + 1);
      root = merg(a, merg(b, c));
    }
    cout << C(res, n) << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    make();
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  