#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    map<int, vector<int> > mp;
    set<int> unvisited_intervals;
    vector<pair<int, int> > all_intervals;
    for (int i = 1; i <= n; ++i) {
      int l, r;
      cin >> l >> r;
      all_intervals.push_back({l, r});
      mp[l].push_back(i);
      mp[r].push_back(-i);
      unvisited_intervals.insert(i);
    }
    sort(all_intervals.begin(), all_intervals.end());
    bool intersection = false;
    for (int i = 0; i < n - 1; ++i) {
      if (all_intervals[i].second >= all_intervals[i + 1].first) {
        intersection = true;
        break;
      }
    }
    if (!intersection) {
      cout << n - 1 << endl;
      continue;
    }
    vector<int> cnt(n + 1, 0), status(n + 1, 0);
    set<int> interval_idx;
    int num_components = 0;
    for (auto &item : mp) {
      int sz = item.second.size(), interval_OI = -1;
      bool some_intervals_erased = false;
      bool some_intervals_added = false;
      if (interval_idx.size() == 1) {
        interval_OI = *interval_idx.begin();
      }
      for (int idx : item.second) {
        if (idx > 0) {
          interval_idx.insert(idx);
          some_intervals_added = true;
        } else {
          interval_idx.erase(-idx);
          some_intervals_erased = true;
          if (unvisited_intervals.find(-idx) != unvisited_intervals.end()) {
            unvisited_intervals.erase(-idx);
          }
        }
      }
      if (some_intervals_added && interval_OI != -1 && status[interval_OI]) {
        ++cnt[interval_OI];
        status[interval_OI] = 0;
      }
      int depth = interval_idx.size();
      num_components += (depth == 0);
      if (depth == 1 && some_intervals_erased) {
        status[*interval_idx.begin()] = 1;
      }
    }
    int max_cnt = -1;
    for (int i = 1; i <= n; ++i) {
      max_cnt = max(max_cnt, cnt[i]);
    }
    cout << num_components + max_cnt << endl;
  }
  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>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    map<int, vector<int> > mp;
    set<int> unvisited_intervals;
    vector<pair<int, int> > all_intervals;
    for (int i = 1; i <= n; ++i) {
      int l, r;
      cin >> l >> r;
      all_intervals.push_back({l, r});
      mp[l].push_back(i);
      mp[r].push_back(-i);
      unvisited_intervals.insert(i);
    }
    sort(all_intervals.begin(), all_intervals.end());
    bool intersection = false;
    for (int i = 0; i < n - 1; ++i) {
      if (all_intervals[i].second >= all_intervals[i + 1].first) {
        intersection = true;
        break;
      }
    }
    if (!intersection) {
      cout << n - 1 << endl;
      continue;
    }
    vector<int> cnt(n + 1, 0), status(n + 1, 0);
    set<int> interval_idx;
    int num_components = 0;
    for (auto &item : mp) {
      int sz = item.second.size(), interval_OI = -1;
      bool some_intervals_erased = false;
      bool some_intervals_added = false;
      if (interval_idx.size() == 1) {
        interval_OI = *interval_idx.begin();
      }
      for (int idx : item.second) {
        if (idx > 0) {
          interval_idx.insert(idx);
          some_intervals_added = true;
        } else {
          interval_idx.erase(-idx);
          some_intervals_erased = true;
          if (unvisited_intervals.find(-idx) != unvisited_intervals.end()) {
            unvisited_intervals.erase(-idx);
          }
        }
      }
      if (some_intervals_added && interval_OI != -1 && status[interval_OI]) {
        ++cnt[interval_OI];
        status[interval_OI] = 0;
      }
      int depth = interval_idx.size();
      num_components += (depth == 0);
      if (depth == 1 && some_intervals_erased) {
        status[*interval_idx.begin()] = 1;
      }
    }
    int max_cnt = -1;
    for (int i = 1; i <= n; ++i) {
      max_cnt = max(max_cnt, cnt[i]);
    }
    cout << num_components + max_cnt << endl;
  }
  return 0;
}

```