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 bool debug = 0;
  const int inf = 1e9;
  mt19937 rng(199999999973);
  struct treap {
    pair<int, int> line;
    int pr;
    int sz;
    treap *l, *r;
    treap(pair<int, int> _line = {-1, -1}) {
      pr = rng();
      line = _line;
      sz = line.second - line.first + 1;
      l = r = nullptr;
    }
  };
  vector<pair<int, int> > travel;
  void dfs(treap* root) {
    if (root == nullptr) return;
    dfs(root->l), travel.push_back(root->line), dfs(root->r);
  }
  typedef pair<treap*, treap*> pT;
  treap* root = nullptr;
  int getsz(treap* v) { return (v ? v->sz : 0); }
  void upd(treap* v) {
    if (v) v->sz = v->line.second - v->line.first + 1 + getsz(v->l) + getsz(v->r);
  }
  treap* merge(treap* l, treap* r) {
    if (l == nullptr) return r;
    if (r == nullptr) return l;
    if (l->pr > r->pr) {
      l->r = merge(l->r, r);
      upd(l);
      return l;
    } else {
      r->l = merge(l, r->l);
      upd(r);
      return r;
    }
  }
  pT split(treap* t, int sz) {
    if (t == nullptr) return {nullptr, nullptr};
    if (getsz(t->l) >= sz) {
      pT left = split(t->l, sz);
      t->l = left.second;
      upd(t);
      return {left.first, t};
    } else {
      if (getsz(t) - getsz(t->r) > sz) {
        treap* v = t->l;
        t->l = nullptr;
        upd(t);
        return {v, t};
      }
      pT right = split(t->r, sz - (getsz(t) - getsz(t->r)));
      t->r = right.first;
      upd(t);
      return {t, right.second};
    }
  }
  treap* add(treap* x, int pos, pair<int, int> line) {
    pT fir = split(x, pos);
    treap* v = new treap(line);
    int ps = getsz(fir.first);
    assert(ps <= pos);
    if (ps < pos) {
      treap* z = fir.second;
      treap* pr = nullptr;
      while (z->l) {
        pr = z, z = z->l;
      }
      if (pr) pr->l = nullptr;
      pair<int, int> kk = z->line;
      pair<int, int> KK;
      if (kk.second - kk.first + 1 <= pos - ps) assert(false);
      KK.first = kk.first, KK.second = kk.first + (pos - ps) - 1;
      kk.first = KK.second + 1;
      assert(kk.first <= kk.second);
      treap* v1 = new treap(KK);
      if (pr) {
        z->line = kk;
        upd(z);
        root = merge(fir.first, v1);
        root = merge(root, v);
        root = merge(root, z);
        root = merge(root, fir.second);
      } else {
        z->line = kk;
        upd(z);
        root = merge(fir.first, v1);
        root = merge(root, v);
        root = merge(root, fir.second);
      }
    } else {
      root = merge(merge(fir.first, v), fir.second);
    }
    return root;
  }
  struct comb {
    static const int COMBN = 4e5 + 10;
    static const int mod = 998244353;
    int fact[COMBN], inv[COMBN];
    int binpow(int a, int b) {
      if (b == 0) return 1;
      int ans = binpow(a, b >> 1);
      ans = (1ll * ans * ans) % mod;
      if (b & 1) ans = (1ll * ans * a) % mod;
      return ans;
    }
    void init() {
      fact[0] = 1;
      for (int i = 1; i < COMBN; ++i) fact[i] = (1ll * fact[i - 1] * i) % mod;
      inv[COMBN - 1] = binpow(fact[COMBN - 1], mod - 2);
      for (int i = COMBN - 2; i >= 0; --i) {
        inv[i] = (1ll * inv[i + 1] * (i + 1)) % mod;
      }
    }
    int Cnk(int n, int k) {
      return (1ll * fact[n] * inv[k] % mod * inv[n - k] % mod);
    }
  } COMB;
  void solve(int n, int m, vector<pair<int, int> > lol) {
    root = new treap({1, 1});
    int lst = 1;
    for (int f = 0; f < (int)(lol).size(); ++f) {
      if (lol[f].first - 1 > lst) {
        root = add(root, lst, make_pair(lst + 1, lol[f].first - 1));
      }
      root = add(root, lol[f].second - 1, {lol[f].first, lol[f].first});
      lst = lol[f].first;
    }
    if (lst < n) root = add(root, lst, make_pair(lst + 1, n));
    travel.clear();
    dfs(root);
    int cnt = n - 1;
    for (int f = 1; f < (int)(travel).size(); ++f) {
      if (travel[f - 1].second > travel[f].first) cnt--;
    }
    cout << COMB.Cnk(n + cnt, cnt) << '\n';
  }
  signed main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int t;
    cin >> t;
    COMB.init();
    while (t--) {
      vector<pair<int, int> > lol;
      int n, m;
      if (!debug) {
        cin >> n >> m;
        for (int i = 0; i < m; ++i) {
          int x, y;
          cin >> x >> y;
          lol.push_back({x, y});
        }
      } else {
        n = 10, m = rng() % 10;
        vector<int> perm(n);
        iota((perm).begin(), (perm).end(), 1);
        shuffle((perm).begin(), (perm).end(), rng);
        for (int f = 0; f < m; ++f) {
          if (perm[f] != 1) lol.push_back({perm[f], rng() % (perm[f] - 1) + 1});
        }
        sort((lol).begin(), (lol).end());
        m = (int)(lol).size();
        cout << "YO\n";
        cout << n << ' ' << m << endl;
        for (auto u : lol) cout << u.first << ' ' << u.second << endl;
        cout << endl;
      }
      solve(n, m, lol);
    }
    return 0;
  }
  