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 = (int)1e9;
  const long long INF = (long long)5e18;
  const int MOD = 998244353;
  int _abs(int x) { return x < 0 ? -x : x; }
  int add(int x, int y) {
    x += y;
    return x >= MOD ? x - MOD : x;
  }
  int sub(int x, int y) {
    x -= y;
    return x < 0 ? x + MOD : x;
  }
  void Add(int &x, int y) {
    x += y;
    if (x >= MOD) x -= MOD;
  }
  void Sub(int &x, int y) {
    x -= y;
    if (x < 0) x += MOD;
  }
  void Mul(int &x, int y) { x = (long long)(x) * (y) % MOD; }
  int qpow(int x, int y) {
    int ret = 1;
    while (y) {
      if (y & 1) ret = (long long)(ret) * (x) % MOD;
      x = (long long)(x) * (x) % MOD;
      y >>= 1;
    }
    return ret;
  }
  void checkmin(int &x, int y) {
    if (x > y) x = y;
  }
  void checkmax(int &x, int y) {
    if (x < y) x = y;
  }
  void checkmin(long long &x, long long y) {
    if (x > y) x = y;
  }
  void checkmax(long long &x, long long y) {
    if (x < y) x = y;
  }
  inline int read() {
    int x = 0, f = 1;
    char c = getchar();
    while (c > '9' || c < '0') {
      if (c == '-') f = -1;
      c = getchar();
    }
    while (c >= '0' && c <= '9')
      x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return x * f;
  }
  const int N = 600005;
  int fac[N], ifac[N];
  int C(int n, int m) {
    return (long long)(fac[n]) * ((long long)(ifac[m]) * (ifac[n - m]) % MOD) %
           MOD;
  }
  void prepare(int n) {
    fac[0] = 1;
    for (int i = 1; i <= n; i++) fac[i] = (long long)(fac[i - 1]) * (i) % MOD;
    ifac[n] = qpow(fac[n], MOD - 2);
    for (int i = n - 1; i >= 0; i--)
      ifac[i] = (long long)(ifac[i + 1]) * (i + 1) % MOD;
  }
  int n, m;
  int root = 0, tot = 0, a[N], id[N];
  struct Tree {
    int ch[2], sz, val, fa;
  } tree[N];
  bool check(int x) { return tree[tree[x].fa].ch[1] == x; }
  inline void pushup(int x) {
    int ls = tree[x].ch[0];
    int rs = tree[x].ch[1];
    tree[x].sz = tree[ls].sz + tree[rs].sz + 1;
  }
  inline void rotate(int x) {
    int y = tree[x].fa, z = tree[y].fa, k = check(x), w = tree[x].ch[k ^ 1];
    tree[z].ch[check(y)] = x;
    tree[x].fa = z;
    tree[x].ch[k ^ 1] = y;
    tree[y].fa = x;
    tree[y].ch[k] = w;
    tree[w].fa = y;
    pushup(y);
    pushup(x);
  }
  void splay(int x, int goal = 0) {
    while (tree[x].fa != goal) {
      int y = tree[x].fa, z = tree[y].fa;
      if (z != goal) {
        if (check(x) == check(y))
          rotate(y);
        else
          rotate(x);
      }
      rotate(x);
    }
    if (!goal) root = x;
  }
  int hcy = 0;
  int kth(int k, int now) {
    hcy++;
    int ls = tree[now].ch[0];
    if (ls == 0) tree[ls].sz = 0;
    if (tree[ls].sz + 1 == k)
      return now;
    else if (tree[ls].sz + 1 < k)
      return kth(k - tree[ls].sz - 1, tree[now].ch[1]);
    else
      return kth(k, tree[now].ch[0]);
  }
  void build(int l, int r, int f) {
    int mid = (l + r) >> 1, now = mid;
    if (l == r) {
      tree[now].sz = 1;
      tree[now].val = a[mid];
    }
    if (l < mid) build(l, mid - 1, mid);
    if (r > mid) build(mid + 1, r, mid);
    tree[now].val = a[mid];
    tree[now].fa = f;
    pushup(now);
    tree[f].ch[mid >= f] = now;
  }
  void insert(int k, int val) {
    int Rt = ++tot;
    int x = kth(k, root);
    int y = kth(k + 1, root);
    splay(x);
    splay(y, x);
    tree[Rt].fa = y;
    tree[y].ch[0] = Rt;
    tree[Rt].val = val;
    tree[Rt].sz = 1;
    pushup(y);
    pushup(x);
    splay(root);
  }
  void del(int k) {
    int x = kth(k - 1, root);
    int y = kth(k + 1, root);
    splay(x);
    splay(y, x);
    int z = tree[y].ch[0];
    tree[y].ch[0] = 0;
    pushup(y);
    pushup(x);
  }
  int Val(int x) {
    int kk = kth(x + 1, root);
    return tree[kk].val;
    splay(kk);
  }
  int x[N], y[N];
  void solve() {
    n = read();
    m = read();
    int ans = n - 1;
    for (int i = 1; i <= m; i++) {
      x[i] = read(), y[i] = read();
      if (y[i] == 1)
        ans--;
      else {
        int num1 = Val(y[i] - 1), num2 = Val(y[i]);
        ans -= (num1 <= num2);
      }
      int num = Val(x[i]);
      del(x[i] + 1);
      insert(y[i], num);
    }
    printf("%d\n", C(n + ans, n));
    for (int i = m; i >= 1; i--) {
      int num = Val(y[i]);
      del(y[i] + 1);
      insert(x[i], num);
    }
  }
  int main() {
    prepare(400000);
    tot = 2;
    a[1] = 0;
    a[2] = inf;
    build(1, 2, 0);
    root = 1;
    tree[0].sz = 0;
    for (int i = 1; i <= 200000; i++) insert(i, i);
    int T = read();
    while (T--) {
      solve();
    }
    cerr << "hcy" << '=' << hcy << endl;
    return 0;
  }
  