#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
const long long inf64 = 1e18;
const int mod = 998244353;
inline int _sum(int a, int b) { return a + b < mod ? a + b : a + b - mod; }
inline int _mul(int a, int b) { return (1ll * a * b) % mod; }
inline int _sub(int a, int b) { return a >= b ? a - b : a - b + mod; }
inline void _inc(int& x, int y) {
  if ((x += y) >= mod) {
    x -= mod;
  }
}
inline void _dec(int& x, int y) {
  if ((x -= y) < 0) {
    x += mod;
  }
}
inline int _binpow(int x, int pw) {
  int res = 1;
  int tmp = x;
  while (pw > 0) {
    if (pw & 1) {
      res = _mul(res, tmp);
    }
    tmp = _mul(tmp, tmp);
    pw >>= 1;
  }
  return res;
}
inline int _div(int a, int b) { return _mul(a, _binpow(b, mod - 2)); }
int slow(int n, int m, vector<pair<int, int> > a) {
  int res = 0;
  vector<int> p(n + 1);
  auto check = [&]() -> int {
    vector<int> pp = p;
    vector<pair<int, int> > b;
    for (int i = 2; i <= n; i++) {
      if (p[i] >= p[i - 1]) continue;
      int j = i - 1;
      while (j > 1 && p[j - 1] > p[i]) j--;
      int value = p[i];
      for (int q = i; q > j; q--) p[q] = p[q - 1];
      p[j] = value;
      b.emplace_back(i, j);
    }
    p = pp;
    return int(a == b);
  };
  function<void(int)> rec = [&](int i) {
    if (i > n) {
      _inc(res, check());
      return;
    }
    for (int x = 1; x <= n; x++) {
      p[i] = x;
      rec(i + 1);
    }
  };
  rec(1);
  return res;
}
mt19937 gen(42);
struct Treap {
  struct Node {
    int prior = gen();
    Node* l = nullptr;
    Node* r = nullptr;
    int value = 0;
    int pu = 0;
    Node(int x) { value = x; }
  };
  void upd(Node* v, int x) {
    if (!v) return;
    v->value += x;
    v->pu += x;
  }
  void push(Node* v) {
    if (!v) return;
    upd(v->l, v->pu);
    upd(v->r, v->pu);
    v->pu = 0;
  }
  void merge(Node*& v, Node* l, Node* r) {
    if (!l || !r) {
      v = l ? l : r;
      return;
    }
    push(l);
    push(r);
    if (l->prior < r->prior) {
      merge(l->r, l->r, r);
      v = l;
    } else {
      merge(r->l, l, r->l);
      v = r;
    }
  }
  void split_by_value(Node* v, Node*& l, Node*& r, int x) {
    if (!v) {
      l = r = nullptr;
      return;
    }
    push(v);
    if (v->value >= x) {
      split_by_value(v->l, l, v->l, x);
      r = v;
    } else {
      split_by_value(v->r, v->r, r, x);
      l = v;
    }
  }
  Node* root = nullptr;
  Node *nl, *nm, *nr;
  Treap() = default;
  void insert(int x) {
    nl = nm = nr = nullptr;
    split_by_value(root, nl, nr, x);
    split_by_value(nr, nm, nr, x + 1);
    if (!nm) nm = new Node(x);
    merge(nl, nl, nm);
    merge(root, nl, nr);
  }
  void erase(int x) {
    nl = nm = nr = nullptr;
    split_by_value(root, nl, nr, x);
    split_by_value(nr, nm, nr, x + 1);
    merge(root, nl, nr);
  }
  void inc(int x) {
    nl = nr = nullptr;
    split_by_value(root, nl, nr, x);
    upd(nr, +1);
    merge(root, nl, nr);
  }
  int dfs(Node* v) {
    if (!v) return 0;
    return dfs(v->l) + dfs(v->r) + 1;
  }
  int size() { return dfs(root); }
};
const int N = 4e5 + 100;
int fact[N];
int rfact[N];
void precalc() {
  fact[0] = 1;
  for (int i = 1; i < N; i++) fact[i] = _mul(fact[i - 1], i);
  rfact[N - 1] = _binpow(fact[N - 1], mod - 2);
  for (int i = N - 2; i >= 0; i--) rfact[i] = _mul(rfact[i + 1], i + 1);
}
int CNK(int n, int k) {
  if (n < k) return 0;
  return _mul(fact[n], _mul(rfact[k], rfact[n - k]));
}
int fast(int n, int m, vector<pair<int, int> > a) {
  Treap tr;
  for (auto [x, y] : a) {
    tr.inc(y);
    tr.insert(y);
    tr.erase(y - 1);
  }
  int sm = tr.size();
  int ub = n - 1 - sm;
  if (ub < 0) return 0;
  return CNK(ub + n, n);
}
void test() {
  mt19937 rnd(42);
  const int MAXN = 6;
  while (1) {
    int n = rnd() % MAXN + 2;
    int m = rnd() % n;
    vector<int> a(n - 1);
    for (int i = 0; i < n - 1; i++) a[i] = i + 2;
    shuffle(a.begin(), a.end(), rnd);
    a.resize(m);
    sort(a.begin(), a.end());
    vector<pair<int, int> > b(m);
    for (int i = 0; i < m; i++)
      b[i].first = a[i], b[i].second = rnd() % (a[i] - 1) + 1;
    int fs = fast(n, m, b);
    int sl = slow(n, m, b);
    if (fs == sl) {
      cout << "OK(" << fs << ")" << endl;
    } else {
      cout << "WA\n";
      cout << "exp = " << sl << "\n";
      cout << "fnd = " << fs << "\n";
      cout << n << " " << m << "\n";
      for (auto [x, y] : b) cout << x << " " << y << "\n";
      break;
    }
  }
}
void work() {
  int t;
  cin >> t;
  while (t--) {
    int n, m;
    cin >> n >> m;
    vector<pair<int, int> > a(m);
    for (auto& [x, y] : a) cin >> x >> y;
    int res = fast(n, m, a);
    cout << res << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  precalc();
  work();
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
const long long inf64 = 1e18;
const int mod = 998244353;
inline int _sum(int a, int b) { return a + b < mod ? a + b : a + b - mod; }
inline int _mul(int a, int b) { return (1ll * a * b) % mod; }
inline int _sub(int a, int b) { return a >= b ? a - b : a - b + mod; }
inline void _inc(int& x, int y) {
  if ((x += y) >= mod) {
    x -= mod;
  }
}
inline void _dec(int& x, int y) {
  if ((x -= y) < 0) {
    x += mod;
  }
}
inline int _binpow(int x, int pw) {
  int res = 1;
  int tmp = x;
  while (pw > 0) {
    if (pw & 1) {
      res = _mul(res, tmp);
    }
    tmp = _mul(tmp, tmp);
    pw >>= 1;
  }
  return res;
}
inline int _div(int a, int b) { return _mul(a, _binpow(b, mod - 2)); }
int slow(int n, int m, vector<pair<int, int> > a) {
  int res = 0;
  vector<int> p(n + 1);
  auto check = [&]() -> int {
    vector<int> pp = p;
    vector<pair<int, int> > b;
    for (int i = 2; i <= n; i++) {
      if (p[i] >= p[i - 1]) continue;
      int j = i - 1;
      while (j > 1 && p[j - 1] > p[i]) j--;
      int value = p[i];
      for (int q = i; q > j; q--) p[q] = p[q - 1];
      p[j] = value;
      b.emplace_back(i, j);
    }
    p = pp;
    return int(a == b);
  };
  function<void(int)> rec = [&](int i) {
    if (i > n) {
      _inc(res, check());
      return;
    }
    for (int x = 1; x <= n; x++) {
      p[i] = x;
      rec(i + 1);
    }
  };
  rec(1);
  return res;
}
mt19937 gen(42);
struct Treap {
  struct Node {
    int prior = gen();
    Node* l = nullptr;
    Node* r = nullptr;
    int value = 0;
    int pu = 0;
    Node(int x) { value = x; }
  };
  void upd(Node* v, int x) {
    if (!v) return;
    v->value += x;
    v->pu += x;
  }
  void push(Node* v) {
    if (!v) return;
    upd(v->l, v->pu);
    upd(v->r, v->pu);
    v->pu = 0;
  }
  void merge(Node*& v, Node* l, Node* r) {
    if (!l || !r) {
      v = l ? l : r;
      return;
    }
    push(l);
    push(r);
    if (l->prior < r->prior) {
      merge(l->r, l->r, r);
      v = l;
    } else {
      merge(r->l, l, r->l);
      v = r;
    }
  }
  void split_by_value(Node* v, Node*& l, Node*& r, int x) {
    if (!v) {
      l = r = nullptr;
      return;
    }
    push(v);
    if (v->value >= x) {
      split_by_value(v->l, l, v->l, x);
      r = v;
    } else {
      split_by_value(v->r, v->r, r, x);
      l = v;
    }
  }
  Node* root = nullptr;
  Node *nl, *nm, *nr;
  Treap() = default;
  void insert(int x) {
    nl = nm = nr = nullptr;
    split_by_value(root, nl, nr, x);
    split_by_value(nr, nm, nr, x + 1);
    if (!nm) nm = new Node(x);
    merge(nl, nl, nm);
    merge(root, nl, nr);
  }
  void erase(int x) {
    nl = nm = nr = nullptr;
    split_by_value(root, nl, nr, x);
    split_by_value(nr, nm, nr, x + 1);
    merge(root, nl, nr);
  }
  void inc(int x) {
    nl = nr = nullptr;
    split_by_value(root, nl, nr, x);
    upd(nr, +1);
    merge(root, nl, nr);
  }
  int dfs(Node* v) {
    if (!v) return 0;
    return dfs(v->l) + dfs(v->r) + 1;
  }
  int size() { return dfs(root); }
};
const int N = 4e5 + 100;
int fact[N];
int rfact[N];
void precalc() {
  fact[0] = 1;
  for (int i = 1; i < N; i++) fact[i] = _mul(fact[i - 1], i);
  rfact[N - 1] = _binpow(fact[N - 1], mod - 2);
  for (int i = N - 2; i >= 0; i--) rfact[i] = _mul(rfact[i + 1], i + 1);
}
int CNK(int n, int k) {
  if (n < k) return 0;
  return _mul(fact[n], _mul(rfact[k], rfact[n - k]));
}
int fast(int n, int m, vector<pair<int, int> > a) {
  Treap tr;
  for (auto [x, y] : a) {
    tr.inc(y);
    tr.insert(y);
    tr.erase(y - 1);
  }
  int sm = tr.size();
  int ub = n - 1 - sm;
  if (ub < 0) return 0;
  return CNK(ub + n, n);
}
void test() {
  mt19937 rnd(42);
  const int MAXN = 6;
  while (1) {
    int n = rnd() % MAXN + 2;
    int m = rnd() % n;
    vector<int> a(n - 1);
    for (int i = 0; i < n - 1; i++) a[i] = i + 2;
    shuffle(a.begin(), a.end(), rnd);
    a.resize(m);
    sort(a.begin(), a.end());
    vector<pair<int, int> > b(m);
    for (int i = 0; i < m; i++)
      b[i].first = a[i], b[i].second = rnd() % (a[i] - 1) + 1;
    int fs = fast(n, m, b);
    int sl = slow(n, m, b);
    if (fs == sl) {
      cout << "OK(" << fs << ")" << endl;
    } else {
      cout << "WA\n";
      cout << "exp = " << sl << "\n";
      cout << "fnd = " << fs << "\n";
      cout << n << " " << m << "\n";
      for (auto [x, y] : b) cout << x << " " << y << "\n";
      break;
    }
  }
}
void work() {
  int t;
  cin >> t;
  while (t--) {
    int n, m;
    cin >> n >> m;
    vector<pair<int, int> > a(m);
    for (auto& [x, y] : a) cin >> x >> y;
    int res = fast(n, m, a);
    cout << res << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  precalc();
  work();
  return 0;
}

```