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;
  const int precLen = 400000;
  long long qpow(long long a, long long n) {
    long long res = 1;
    while (n) {
      if (n & 1) res = res * a % mod;
      a = a * a % mod;
      n >>= 1;
    }
    return res;
  }
  int fact[precLen + 5], inv[precLen + 5];
  long long C(int n, int r) {
    return (long long)fact[n] * inv[n - r] % mod * inv[r] % mod;
  }
  void initialization() {
    fact[0] = 1;
    for (int i = 1; i <= precLen; i++) {
      fact[i] = (long long)fact[i - 1] * i % mod;
    }
    inv[precLen] = qpow(fact[precLen], mod - 2);
    for (int i = precLen - 1; i >= 0; i--) {
      inv[i] = (long long)inv[i + 1] * (i + 1) % mod;
    }
  }
  mt19937 TreapRnd(chrono::steady_clock().now().time_since_epoch().count());
  struct Treap {
    struct node {
      node *l, *r;
      int val, pri, sz, laz, lvl;
      node() {
        l = r = nullptr;
        val = laz = lvl = 0;
        sz = 1;
        pri = TreapRnd();
      }
      void pullup() {
        sz = (l == nullptr ? 0 : l->sz) + 1 + (r == nullptr ? 0 : r->sz);
      }
      void pushdown() {
        if (l != nullptr) {
          l->laz += laz;
          l->val += laz;
        }
        if (r != nullptr) {
          r->laz += laz;
          r->val += laz;
        }
        laz = 0;
      }
    };
    int size(node *p) {
      if (p == nullptr) return 0;
      return p->sz;
    }
    static int poolsz;
    static node *pool;
    node *getnode() {
      if (poolsz == 0) {
        pool = new node[poolsz = 114514];
      }
      poolsz--;
      return pool++;
    }
    node *root;
    Treap() { root = nullptr; }
    void merge(node *&p, node *l, node *r) {
      if (l == nullptr) {
        p = r;
      } else if (r == nullptr) {
        p = l;
      } else if (l->pri < r->pri) {
        l->pushdown();
        merge(l->r, l->r, r);
        l->pullup();
        p = l;
      } else {
        r->pushdown();
        merge(r->l, l, r->l);
        r->pullup();
        p = r;
      }
    }
    void split(node *p, int k, node *&l, node *&r) {
      if (p == nullptr) {
        l = r = nullptr;
      } else if (p->val > k) {
        p->pushdown();
        split(p->l, k, l, p->l);
        p->pullup();
        r = p;
      } else {
        p->pushdown();
        split(p->r, k, p->r, r);
        p->pullup();
        l = p;
      }
    }
    void ins(int x, int y) {
      node *l, *r, *p;
      split(root, x, l, r);
      split(l, x - 1, l, p);
      p = getnode();
      p->val = x;
      if (y) p->lvl = y;
      merge(l, l, p);
      merge(root, l, r);
    }
    int fnd(int x) {
      node *l, *r, *p;
      split(root, x, l, r);
      split(l, x - 1, l, p);
      int res = p != nullptr ? p->lvl : 0;
      merge(l, l, p);
      merge(root, l, r);
      return res;
    }
    void add(int x) {
      node *l, *r;
      split(root, x - 1, l, r);
      if (r != nullptr) {
        r->val++;
        r->laz++;
      }
      merge(root, l, r);
    }
  };
  int Treap::poolsz = 0;
  Treap::node *Treap::pool = nullptr;
  void mian() {
    int n, m;
    cin >> n >> m;
    Treap t;
    int cnt = 0;
    for (int i = 0; i < m; i++) {
      int x, y;
      cin >> x >> y;
      if (t.fnd(y - 1) == 1) {
        t.add(y);
        t.ins(y, 1);
        t.ins(y - 1, 2);
      } else {
        t.add(y);
        t.ins(y, 1);
        cnt++;
      }
    }
    cout << C(n + n - cnt - 1, n - cnt - 1) << '\n';
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    initialization();
    int T;
    cin >> T;
    while (T--) mian();
    return 0;
  }
  