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;
  struct BIT {
    vector<long long> bit;
    long long n;
    BIT(long long size) {
      n = size + 1;
      bit.assign(n, 0);
    }
    void upd(long long pos, long long val) {
      for (++pos; pos < n; pos += pos & -pos) {
        bit[pos] += val;
      }
    }
    long long get(long long pos) {
      long long ret = 0;
      for (++pos; pos > 0; pos -= pos & -pos) {
        ret += bit[pos];
      }
      return ret;
    }
  };
  const long long mod = 998244353;
  long long x[200005], y[200005];
  long long pos[200005];
  long long binpow(long long a, long long b) {
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % mod;
      a = a * a % mod;
      b >>= 1;
    }
    return res;
  }
  long long fact[2 * 200005];
  long long ans(long long a, long long b) {
    return fact[a] * binpow(fact[b], mod - 2) % mod *
           binpow(fact[a - b], mod - 2) % mod;
  }
  BIT bt(200005);
  void solve() {
    long long n, m;
    cin >> n >> m;
    for (long long i = 1; i <= m; i++) {
      cin >> x[i] >> y[i];
    }
    map<long long, long long> ck;
    for (long long i = m; i >= 1; i--) {
      long long l = 1, r = n;
      while (l < r) {
        long long m = (l + r) / 2;
        if (m - bt.get(m) < y[i]) {
          l = m + 1;
        } else
          r = m;
      }
      pos[i] = l;
      ck[l] = 1;
      bt.upd(l, 1);
    }
    set<long long> s2;
    for (long long i = 1; i <= m; i++) {
      if (!ck[pos[i] + 1]) {
        s2.insert(pos[i] + 1);
      }
    }
    map<long long, long long> mp;
    long long cnt = 0;
    for (long long i = 1; i <= m; i++) {
      long long mini = 1e9;
      if (s2.size() && *s2.rbegin() > pos[i]) {
        mini = min(mini, *s2.upper_bound(pos[i]));
      }
      if (!mp[mini]) cnt++;
      mp[mini] = 1;
      s2.insert(pos[i]);
    }
    for (long long i = 1; i <= m; i++) {
      bt.upd(pos[i], -1);
    }
    cnt = n - 1 - cnt;
    cout << ans(n + cnt, cnt) << endl;
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    fact[0] = 1;
    for (long long i = 1; i < 2 * 200005; i++) {
      fact[i] = fact[i - 1] * i % mod;
    }
    long long t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  