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>
  #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
  using namespace std;
  template <class T>
  using rque = priority_queue<T, vector<T>, greater<T>>;
  template <class T>
  bool chmin(T &a, const T &b) {
    if (b < a) {
      a = b;
      return 1;
    }
    return 0;
  }
  template <class T>
  bool chmax(T &a, const T &b) {
    if (b > a) {
      a = b;
      return 1;
    }
    return 0;
  }
  long long gcd(long long a, long long b) {
    if (a == 0) return b;
    if (b == 0) return a;
    long long cnt = a % b;
    while (cnt != 0) {
      a = b;
      b = cnt;
      cnt = a % b;
    }
    return b;
  }
  long long extGCD(long long a, long long b, long long &x, long long &y) {
    if (b == 0) {
      x = 1;
      y = 0;
      return a;
    }
    long long d = extGCD(b, a % b, y, x);
    y -= a / b * x;
    return d;
  }
  struct UnionFind {
    vector<long long> data;
    int num;
    UnionFind(int sz) {
      data.assign(sz, -1);
      num = sz;
    }
    bool unite(int x, int y) {
      x = find(x), y = find(y);
      if (x == y) return (false);
      if (data[x] > data[y]) swap(x, y);
      data[x] += data[y];
      data[y] = x;
      num--;
      return (true);
    }
    int find(int k) {
      if (data[k] < 0) return (k);
      return (data[k] = find(data[k]));
    }
    long long size(int k) { return (-data[find(k)]); }
    bool same(int x, int y) { return find(x) == find(y); }
  };
  template <int mod>
  struct ModInt {
    int x;
    ModInt() : x(0) {}
    ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
    ModInt &operator+=(const ModInt &p) {
      if ((x += p.x) >= mod) x -= mod;
      return *this;
    }
    ModInt &operator-=(const ModInt &p) {
      if ((x += mod - p.x) >= mod) x -= mod;
      return *this;
    }
    ModInt &operator*=(const ModInt &p) {
      x = (int)(1LL * x * p.x % mod);
      return *this;
    }
    ModInt &operator/=(const ModInt &p) {
      *this *= p.inverse();
      return *this;
    }
    ModInt operator-() const { return ModInt(-x); }
    ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }
    ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }
    ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }
    ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }
    bool operator==(const ModInt &p) const { return x == p.x; }
    bool operator!=(const ModInt &p) const { return x != p.x; }
    ModInt inverse() const {
      int a = x, b = mod, u = 1, v = 0, t;
      while (b > 0) {
        t = a / b;
        swap(a -= t * b, b);
        swap(u -= t * v, v);
      }
      return ModInt(u);
    }
    ModInt pow(int64_t n) const {
      ModInt ret(1), mul(x);
      while (n > 0) {
        if (n & 1) ret *= mul;
        mul *= mul;
        n >>= 1;
      }
      return ret;
    }
    friend ostream &operator<<(ostream &os, const ModInt &p) { return os << p.x; }
    friend istream &operator>>(istream &is, ModInt &a) {
      int64_t t;
      is >> t;
      a = ModInt<mod>(t);
      return (is);
    }
    static int get_mod() { return mod; }
  };
  template <typename Monoid>
  struct SegmentTree {
    using F = function<Monoid(Monoid, Monoid)>;
    int sz;
    vector<Monoid> seg;
    const F f;
    const Monoid M1;
    SegmentTree(int n, const F f, const Monoid &M1) : f(f), M1(M1) {
      sz = 1;
      while (sz < n) sz <<= 1;
      seg.assign(2 * sz, M1);
    }
    void set(int k, const Monoid &x) { seg[k + sz] = x; }
    void build() {
      for (int k = sz - 1; k > 0; k--) {
        seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
      }
    }
    void update(int k, const Monoid &x) {
      k += sz;
      seg[k] = x;
      while (k >>= 1) {
        seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
      }
    }
    Monoid query(int a, int b) {
      Monoid L = M1, R = M1;
      for (a += sz, b += sz; a < b; a >>= 1, b >>= 1) {
        if (a & 1) L = f(L, seg[a++]);
        if (b & 1) R = f(seg[--b], R);
      }
      return f(L, R);
    }
    Monoid operator[](const int &k) const { return seg[k + sz]; }
    template <typename C>
    int find_subtree(int a, const C &check, Monoid &M, bool type) {
      while (a < sz) {
        Monoid nxt = type ? f(seg[2 * a + type], M) : f(M, seg[2 * a + type]);
        if (check(nxt))
          a = 2 * a + type;
        else
          M = nxt, a = 2 * a + 1 - type;
      }
      return a - sz;
    }
    template <typename C>
    int find_first(int a, const C &check) {
      Monoid L = M1;
      if (a <= 0) {
        if (check(f(L, seg[1]))) return find_subtree(1, check, L, false);
        return -1;
      }
      int b = sz;
      for (a += sz, b += sz; a < b; a >>= 1, b >>= 1) {
        if (a & 1) {
          Monoid nxt = f(L, seg[a]);
          if (check(nxt)) return find_subtree(a, check, L, false);
          L = nxt;
          ++a;
        }
      }
      return -1;
    }
    template <typename C>
    int find_last(int b, const C &check) {
      Monoid R = M1;
      if (b >= sz) {
        if (check(f(seg[1], R))) return find_subtree(1, check, R, true);
        return -1;
      }
      int a = sz;
      for (b += sz; a < b; a >>= 1, b >>= 1) {
        if (b & 1) {
          Monoid nxt = f(seg[--b], R);
          if (check(nxt)) return find_subtree(b, check, R, true);
          R = nxt;
        }
      }
      return -1;
    }
  };
  constexpr int mod = 998244353;
  using mint = ModInt<mod>;
  mint mpow(mint x, long long n) {
    mint ans = 1;
    while (n != 0) {
      if (n & 1) ans *= x;
      x *= x;
      n = n >> 1;
    }
    return ans;
  }
  long long mpow2(long long x, long long n, long long mod) {
    long long ans = 1;
    while (n != 0) {
      if (n & 1) ans = ans * x % mod;
      x = x * x % mod;
      n = n >> 1;
    }
    return ans;
  }
  vector<mint> fac;
  vector<mint> ifac;
  void setcomb(int sz = 2000010) {
    fac.assign(sz + 1, 0);
    ifac.assign(sz + 1, 0);
    fac[0] = 1;
    for (long long i = 0; i < sz; i++) {
      fac[i + 1] = fac[i] * (i + 1);
    }
    ifac[sz] = fac[sz].inverse();
    for (long long i = sz; i > 0; i--) {
      ifac[i - 1] = ifac[i] * i;
    }
  }
  mint comb(long long a, long long b) {
    if (fac.size() == 0) setcomb();
    if (a == 0 && b == 0) return 1;
    if (a < b || a < 0 || b < 0) return 0;
    return ifac[a - b] * ifac[b] * fac[a];
  }
  mint perm(long long a, long long b) {
    if (fac.size() == 0) setcomb();
    if (a == 0 && b == 0) return 1;
    if (a < b || a < 0) return 0;
    return fac[a] * ifac[a - b];
  }
  long long modinv2(long long a, long long mod) {
    long long b = mod, u = 1, v = 0;
    while (b) {
      long long t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    u %= mod;
    if (u < 0) u += mod;
    return u;
  }
  int main() {
    ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    cout << fixed << setprecision(15);
    SegmentTree<long long> seg(
        200001, [](long long a, long long b) { return a + b; }, 0);
    for (long long i = 0; i < (200001); i++) seg.set(i, -1);
    seg.build();
    int T;
    cin >> T;
    while (T--) {
      long long n, m;
      cin >> n >> m;
      if (m == 0) {
        cout << comb(2 * n - 1, n - 1) << endl;
        continue;
      }
      long long x[m], y[m];
      for (long long i = 0; i < (m); i++) cin >> x[i] >> y[i], y[i]--;
      seg.update(n, -1e9);
      int flag = 0;
      for (long long i = (long long)m - 1; i >= 0; i--) {
        y[i] = seg.find_first(0, [&](long long x) { return y[i] + x < -1; });
        if (y[i] >= n)
          flag = 1;
        else
          seg.update(y[i], 0);
      }
      for (long long i = 0; i < (m); i++) seg.update(y[i], -1);
      seg.update(n, -1);
      if (flag) {
        cout << 0 << endl;
        continue;
      }
      mint ans = 0;
      long long cnt = 0;
      vector<pair<long long, long long>> v;
      for (long long i = 0; i < (m); i++) v.push_back({y[i], i});
      sort(v.begin(), v.end());
      for (long long i = 0; i < (m - 1); i++)
        if (v[i].first + 1 == v[i + 1].first && v[i].second < v[i + 1].second)
          cnt++;
      for (long long i = 0; i < (cnt + 1); i++)
        ans += comb(2 * n - 1 - m, n - i) * comb(cnt, i);
      cout << ans << endl;
    }
  }
  