#include <bits/stdc++.h>
const int N = 200010;
const long long INF = 1ll << 60;
const int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
class Discretization {
 public:
  Discretization() { vec_.clear(); }
  void Init() { vec_.clear(); }
  void Add(int x) { vec_.push_back(x); }
  void Build() {
    std::sort(vec_.begin(), vec_.end());
    vec_.erase(std::unique(vec_.begin(), vec_.end()), vec_.end());
  }
  int Id(int x) {
    auto y = std::lower_bound(vec_.begin(), vec_.end(), x);
    if (y == vec_.end() || *y != x) {
      return -1;
    } else {
      return y - vec_.begin() + 1;
    }
  }
  int IdLeq(int x) {
    auto y = std::upper_bound(vec_.begin(), vec_.end(), x);
    if (y == vec_.begin()) {
      return 0;
    } else {
      --y;
      return y - vec_.begin() + 1;
    }
  }
  int IdGeq(int x) {
    auto y = std::lower_bound(vec_.begin(), vec_.end(), x);
    if (y == vec_.end()) {
      return vec_.size() + 1;
    } else {
      return y - vec_.begin() + 1;
    }
  }
  int Size() { return vec_.size(); }
  int Value(int id) { return vec_[id - 1]; }

 private:
  std::vector<int> vec_;
} disc;
int t;
int n, m;
int a[N];
int b[N];
std::vector<int> vec[N << 2][2];
std::set<int> se;
std::vector<int> why[N];
int init = 0;
bool left[N], right[N];
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    disc.Init();
    for (int i = 1; i <= n; ++i) {
      scanf("%d%d", &a[i], &b[i]);
      disc.Add(a[i]);
      disc.Add(b[i]);
    }
    disc.Build();
    for (int i = 1; i <= 2 * n; ++i) vec[i][0].clear(), vec[i][1].clear();
    for (int i = 1; i <= n; ++i) {
      vec[disc.Id(a[i])][0].push_back(i);
      vec[disc.Id(b[i])][1].push_back(i);
      why[i].clear();
      left[i] = right[i] = false;
    }
    se.clear();
    int id_before, id_after;
    init = 0;
    for (int i = 1; i <= disc.Size(); ++i) {
      int before = se.size();
      if (before == 1) id_before = *se.begin();
      for (size_t j = 0; j < vec[i][0].size(); ++j) {
        se.insert(vec[i][0][j]);
      }
      int after = se.size();
      if (before == 1 && after > 1) why[id_before].push_back(1);
      if (before == 0 && after == 1) left[*se.begin()] = true;
      before = se.size();
      if (before == 1) id_before = *se.begin();
      for (size_t j = 0; j < vec[i][1].size(); ++j) {
        se.erase(vec[i][1][j]);
      }
      after = se.size();
      if (after == 1) id_after = *se.begin();
      if (before > 1 && after == 1) why[id_after].push_back(-1);
      if (before == 1 && after == 0) right[id_before] = true;
      if (!se.size()) ++init;
    }
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
      if (left[i] && right[i] && !why[i].size()) {
        ans = std::max(ans, init - 1);
      } else {
        int cnt = 0;
        int last = -2;
        for (size_t j = 0; j < why[i].size(); ++j) {
          if (last == -1 && why[i][j] == 1) ++cnt;
          last = why[i][j];
        }
        ans = std::max(ans, init + cnt);
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

There are n segments on a Ox axis [l_1, r_1], [l_2, r_2], ..., [l_n, r_n]. Segment [l, r] covers all points from l to r inclusive, so all x such that l ≤ x ≤ r.

Segments can be placed arbitrarily — be inside each other, coincide and so on. Segments can degenerate into points, that is l_i=r_i is possible.

Union of the set of segments is such a set of segments which covers exactly the same set of points as the original set. For example:

  * if n=3 and there are segments [3, 6], [100, 100], [5, 8] then their union is 2 segments: [3, 8] and [100, 100]; 
  * if n=5 and there are segments [1, 2], [2, 3], [4, 5], [4, 6], [6, 6] then their union is 2 segments: [1, 3] and [4, 6]. 



Obviously, a union is a set of pairwise non-intersecting segments.

You are asked to erase exactly one segment of the given n so that the number of segments in the union of the rest n-1 segments is maximum possible.

For example, if n=4 and there are segments [1, 4], [2, 3], [3, 6], [5, 7], then:

  * erasing the first segment will lead to [2, 3], [3, 6], [5, 7] remaining, which have 1 segment in their union; 
  * erasing the second segment will lead to [1, 4], [3, 6], [5, 7] remaining, which have 1 segment in their union; 
  * erasing the third segment will lead to [1, 4], [2, 3], [5, 7] remaining, which have 2 segments in their union; 
  * erasing the fourth segment will lead to [1, 4], [2, 3], [3, 6] remaining, which have 1 segment in their union. 



Thus, you are required to erase the third segment to get answer 2.

Write a program that will find the maximum number of segments in the union of n-1 segments if you erase any of the given n segments.

Note that if there are multiple equal segments in the given set, then you can erase only one of them anyway. So the set after erasing will have exactly n-1 segments.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test. Then the descriptions of t test cases follow.

The first of each test case contains a single integer n (2 ≤ n ≤ 2⋅10^5) — the number of segments in the given set. Then n lines follow, each contains a description of a segment — a pair of integers l_i, r_i (-10^9 ≤ l_i ≤ r_i ≤ 10^9), where l_i and r_i are the coordinates of the left and right borders of the i-th segment, respectively.

The segments are given in an arbitrary order.

It is guaranteed that the sum of n over all test cases does not exceed 2⋅10^5.

Output

Print t integers — the answers to the t given test cases in the order of input. The answer is the maximum number of segments in the union of n-1 segments if you erase any of the given n segments.

Example

Input


3
4
1 4
2 3
3 6
5 7
3
5 5
5 5
5 5
6
3 3
1 1
5 5
1 5
2 2
4 4


Output


2
1
5

### Response

```cpp
#include <bits/stdc++.h>
const int N = 200010;
const long long INF = 1ll << 60;
const int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
class Discretization {
 public:
  Discretization() { vec_.clear(); }
  void Init() { vec_.clear(); }
  void Add(int x) { vec_.push_back(x); }
  void Build() {
    std::sort(vec_.begin(), vec_.end());
    vec_.erase(std::unique(vec_.begin(), vec_.end()), vec_.end());
  }
  int Id(int x) {
    auto y = std::lower_bound(vec_.begin(), vec_.end(), x);
    if (y == vec_.end() || *y != x) {
      return -1;
    } else {
      return y - vec_.begin() + 1;
    }
  }
  int IdLeq(int x) {
    auto y = std::upper_bound(vec_.begin(), vec_.end(), x);
    if (y == vec_.begin()) {
      return 0;
    } else {
      --y;
      return y - vec_.begin() + 1;
    }
  }
  int IdGeq(int x) {
    auto y = std::lower_bound(vec_.begin(), vec_.end(), x);
    if (y == vec_.end()) {
      return vec_.size() + 1;
    } else {
      return y - vec_.begin() + 1;
    }
  }
  int Size() { return vec_.size(); }
  int Value(int id) { return vec_[id - 1]; }

 private:
  std::vector<int> vec_;
} disc;
int t;
int n, m;
int a[N];
int b[N];
std::vector<int> vec[N << 2][2];
std::set<int> se;
std::vector<int> why[N];
int init = 0;
bool left[N], right[N];
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    disc.Init();
    for (int i = 1; i <= n; ++i) {
      scanf("%d%d", &a[i], &b[i]);
      disc.Add(a[i]);
      disc.Add(b[i]);
    }
    disc.Build();
    for (int i = 1; i <= 2 * n; ++i) vec[i][0].clear(), vec[i][1].clear();
    for (int i = 1; i <= n; ++i) {
      vec[disc.Id(a[i])][0].push_back(i);
      vec[disc.Id(b[i])][1].push_back(i);
      why[i].clear();
      left[i] = right[i] = false;
    }
    se.clear();
    int id_before, id_after;
    init = 0;
    for (int i = 1; i <= disc.Size(); ++i) {
      int before = se.size();
      if (before == 1) id_before = *se.begin();
      for (size_t j = 0; j < vec[i][0].size(); ++j) {
        se.insert(vec[i][0][j]);
      }
      int after = se.size();
      if (before == 1 && after > 1) why[id_before].push_back(1);
      if (before == 0 && after == 1) left[*se.begin()] = true;
      before = se.size();
      if (before == 1) id_before = *se.begin();
      for (size_t j = 0; j < vec[i][1].size(); ++j) {
        se.erase(vec[i][1][j]);
      }
      after = se.size();
      if (after == 1) id_after = *se.begin();
      if (before > 1 && after == 1) why[id_after].push_back(-1);
      if (before == 1 && after == 0) right[id_before] = true;
      if (!se.size()) ++init;
    }
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
      if (left[i] && right[i] && !why[i].size()) {
        ans = std::max(ans, init - 1);
      } else {
        int cnt = 0;
        int last = -2;
        for (size_t j = 0; j < why[i].size(); ++j) {
          if (last == -1 && why[i][j] == 1) ++cnt;
          last = why[i][j];
        }
        ans = std::max(ans, init + cnt);
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}

```