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 INF = 0x3f3f3f3f;
  const int MOD = 998244353;
  const int MAXN = 4e5 + 1;
  int quick(int A, int B) {
    if (B == 0) return 1;
    int tmp = quick(A, B >> 1);
    tmp = 1ll * tmp * tmp % MOD;
    if (B & 1) tmp = 1ll * tmp * A % MOD;
    return tmp;
  }
  int inv(int A) { return quick(A, MOD - 2); }
  int fact[MAXN], ifact[MAXN];
  int comb(int A, int B) {
    return 1ll * fact[A] * ifact[B] % MOD * ifact[A - B] % MOD;
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  vector<pair<int, int> > gt;
  struct Treap {
    int root = 0;
    struct node {
      pair<int, int> val;
      int ran, l, r, siz;
    };
    vector<node> v;
    int new_node(pair<int, int> val) {
      if (v.empty()) {
        v.push_back(node{make_pair(0, -1), (int)rng() % INF, 0, 0, 0});
      }
      v.push_back(node{val, (int)rng() % INF, 0, 0, val.second - val.first + 1});
      return v.size() - 1;
    }
    void update(int index) {
      v[index].siz = v[index].val.second - v[index].val.first + 1;
      if (v[index].l) {
        v[index].siz += v[v[index].l].siz;
      }
      if (v[index].r) {
        v[index].siz += v[v[index].r].siz;
      }
    }
    void split_by_size(int base, int& x, int& y, int boundary) {
      if (!boundary) {
        x = 0;
        y = base;
        return;
      }
      if (!base) {
        x = y = 0;
        return;
      }
      if (v[v[base].l].siz + v[base].val.second - v[base].val.first + 1 <
          boundary) {
        x = base;
        split_by_size(v[base].r, v[x].r, y,
                      boundary - v[v[base].l].siz -
                          (v[base].val.second - v[base].val.first + 1));
        update(x);
      } else {
        y = base;
        split_by_size(v[base].l, x, v[y].l, boundary);
        update(y);
      }
    }
    void getleft(int base, int& x, int& y) {
      if (!base) {
        x = y = 0;
        return;
      }
      if (!v[base].l) {
        x = base;
        y = v[base].r;
        v[base].r = 0;
        update(x);
      } else {
        y = base;
        getleft(v[base].l, x, v[y].l);
        update(y);
      }
    }
    int merge(int x, int y) {
      if (x == 0 || y == 0) return x + y;
      if (v[x].ran > v[y].ran) {
        v[x].r = merge(v[x].r, y);
        update(x);
        return x;
      }
      v[y].l = merge(x, v[y].l);
      update(y);
      return y;
    }
    void traval(int now) {
      if (now == 0) return;
      traval(v[now].l);
      gt.push_back(v[now].val);
      traval(v[now].r);
    }
  } fhq;
  int x[MAXN], y[MAXN];
  void solve() {
    fhq.v.clear();
    fhq.root = 0;
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; ++i) {
      scanf("%d%d", &x[i], &y[i]);
    }
    vector<pair<pair<int, int>, int> > can;
    int pre = 0;
    for (int i = 1; i <= m; ++i) {
      if (pre + 1 <= x[i] - 1) {
        can.push_back(make_pair(make_pair(pre + 1, x[i] - 1), n));
      }
      can.push_back(make_pair(make_pair(x[i], x[i]), y[i]));
      pre = x[i];
    }
    for (auto it : can) {
      if (it.second == n) {
        fhq.root = fhq.merge(fhq.root, fhq.new_node(it.first));
      } else {
        int x, y;
        fhq.split_by_size(fhq.root, x, y, it.second);
        int z;
        fhq.getleft(y, z, y);
        int nw = fhq.new_node(it.first);
        it.second -= fhq.v[x].siz;
        int _, __;
        _ = __ = 0;
        it.second--;
        pair<int, int> orz = fhq.v[z].val;
        if (it.second) {
          _ = fhq.new_node(make_pair(orz.first, orz.first + it.second - 1));
        }
        if (it.second != fhq.v[z].siz) {
          __ = fhq.new_node(make_pair(orz.first + it.second, orz.second));
        }
        fhq.root = fhq.merge(x, fhq.merge(fhq.merge(fhq.merge(_, nw), __), y));
      }
    }
    gt.clear();
    fhq.traval(fhq.root);
    int cnt = 0;
    for (int i = 0; i < gt.size(); ++i) {
      if (i) {
        if (gt[i - 1].second > gt[i].first) {
          cnt++;
        }
      }
    }
    int ans = comb(n - 1 + n - 1 - cnt + 1, n);
    printf("%d\n", ans);
  }
  int main() {
    fact[0] = 1;
    for (int i = 1; i <= 400000; ++i) fact[i] = 1ll * fact[i - 1] * i % MOD;
    for (int i = 0; i <= 400000; ++i) ifact[i] = inv(fact[i]);
    int t;
    scanf("%d", &t);
    while (t--) solve();
    return 0;
  }
  