#include <bits/stdc++.h>
int N;
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  int T;
  std::cin >> T;
  while (T--) {
    std::set<int> smen;
    std::set<int> set;
    std::map<int, std::vector<int>> plus;
    std::map<int, std::vector<int>> minus;
    std::cin >> N;
    for (int i = 0, x, y; i < N; ++i) {
      std::cin >> x >> y;
      smen.insert(x);
      smen.insert(y);
      plus[x].push_back(i);
      minus[y].push_back(i);
    }
    int initial = 0;
    std::vector<int> count(N, 0);
    for (auto &it : smen) {
      for (auto &event : plus[it]) set.insert(event);
      for (auto &event : minus[it]) {
        set.erase(event);
        if (set.size() == 1)
          count[*set.begin()]++;
        else if (set.empty())
          count[event]--;
      }
      initial += (set.empty() ? 1 : 0);
    }
    int ans = count[0];
    for (int i = 1; i < N; ++i) ans = std::max(ans, count[i]);
    ans = ans + initial;
    std::cout << ans << '\n';
  }
  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>
int N;
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  int T;
  std::cin >> T;
  while (T--) {
    std::set<int> smen;
    std::set<int> set;
    std::map<int, std::vector<int>> plus;
    std::map<int, std::vector<int>> minus;
    std::cin >> N;
    for (int i = 0, x, y; i < N; ++i) {
      std::cin >> x >> y;
      smen.insert(x);
      smen.insert(y);
      plus[x].push_back(i);
      minus[y].push_back(i);
    }
    int initial = 0;
    std::vector<int> count(N, 0);
    for (auto &it : smen) {
      for (auto &event : plus[it]) set.insert(event);
      for (auto &event : minus[it]) {
        set.erase(event);
        if (set.size() == 1)
          count[*set.begin()]++;
        else if (set.empty())
          count[event]--;
      }
      initial += (set.empty() ? 1 : 0);
    }
    int ans = count[0];
    for (int i = 1; i < N; ++i) ans = std::max(ans, count[i]);
    ans = ans + initial;
    std::cout << ans << '\n';
  }
  return 0;
}

```