#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int t, n, segcnt;
struct segment {
  int start, end;
  int cnt;
  bool candidate;
  bool isocand;
  bool iso;
} segs[200010];
struct point {
  int x;
  int segid;
  bool endpoint;
  point(int x, int id) : x(x), segid(id) {}
};
vector<point> pt;
bool operator<(const point& p, const point& q) {
  if (p.x != q.x) return p.x < q.x;
  if (!p.endpoint && q.endpoint) return true;
  if (p.endpoint && !q.endpoint) return false;
  return p.segid < q.segid;
  ;
}
void solve() {
  cin >> n;
  pt.clear();
  segcnt = 0;
  for (int i = 0; i < n; ++i) {
    cin >> segs[i].start >> segs[i].end;
    segs[i].cnt = 0;
    segs[i].candidate = false, segs[i].isocand = segs[i].iso = false;
    pt.push_back(point(segs[i].start, i));
    pt.back().endpoint = false;
    pt.push_back(point(segs[i].end, i));
    pt.back().endpoint = true;
  }
  sort(pt.begin(), pt.end());
  set<int> open;
  for (int i = 0; i < (int)pt.size();) {
    if (open.size() == 1 && segs[*open.begin()].candidate) {
      if (!pt[i].endpoint) segs[*open.begin()].cnt++;
      segs[*open.begin()].candidate = false;
    }
    if (open.size() == 1 && segs[*open.begin()].isocand) {
      if (pt[i].endpoint) segs[*open.begin()].iso = true;
      segs[*open.begin()].isocand = false;
    }
    int currx = pt[i].x;
    int stc = open.size();
    while (i < (int)pt.size() && pt[i].x == currx && !pt[i].endpoint) {
      open.insert(pt[i].segid);
      ++i;
    }
    int contain = open.size();
    int idd;
    if (contain == 1) idd = *open.begin();
    while (i < (int)pt.size() && pt[i].x == currx && pt[i].endpoint) {
      open.erase(pt[i].segid);
      ++i;
    }
    if (open.size() == 1 && contain > 1) segs[*open.begin()].candidate = true;
    if (open.size() == 1 && contain == 1 && stc == 0)
      segs[*open.begin()].isocand = true;
    if (open.size() == 0 && contain == 1 && stc == 0) segs[idd].iso = true;
    if (open.size() == 0) ++segcnt;
  }
  int maxc = 0;
  bool onlyiso = true;
  for (int i = 0; i < n; ++i) {
    maxc = max(maxc, segs[i].cnt);
    if (!segs[i].iso) onlyiso = false;
  }
  if (onlyiso)
    cout << segcnt - 1 << endl;
  else
    cout << segcnt + maxc << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Please create a solution in Cpp to 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;
using ll = long long;
int t, n, segcnt;
struct segment {
  int start, end;
  int cnt;
  bool candidate;
  bool isocand;
  bool iso;
} segs[200010];
struct point {
  int x;
  int segid;
  bool endpoint;
  point(int x, int id) : x(x), segid(id) {}
};
vector<point> pt;
bool operator<(const point& p, const point& q) {
  if (p.x != q.x) return p.x < q.x;
  if (!p.endpoint && q.endpoint) return true;
  if (p.endpoint && !q.endpoint) return false;
  return p.segid < q.segid;
  ;
}
void solve() {
  cin >> n;
  pt.clear();
  segcnt = 0;
  for (int i = 0; i < n; ++i) {
    cin >> segs[i].start >> segs[i].end;
    segs[i].cnt = 0;
    segs[i].candidate = false, segs[i].isocand = segs[i].iso = false;
    pt.push_back(point(segs[i].start, i));
    pt.back().endpoint = false;
    pt.push_back(point(segs[i].end, i));
    pt.back().endpoint = true;
  }
  sort(pt.begin(), pt.end());
  set<int> open;
  for (int i = 0; i < (int)pt.size();) {
    if (open.size() == 1 && segs[*open.begin()].candidate) {
      if (!pt[i].endpoint) segs[*open.begin()].cnt++;
      segs[*open.begin()].candidate = false;
    }
    if (open.size() == 1 && segs[*open.begin()].isocand) {
      if (pt[i].endpoint) segs[*open.begin()].iso = true;
      segs[*open.begin()].isocand = false;
    }
    int currx = pt[i].x;
    int stc = open.size();
    while (i < (int)pt.size() && pt[i].x == currx && !pt[i].endpoint) {
      open.insert(pt[i].segid);
      ++i;
    }
    int contain = open.size();
    int idd;
    if (contain == 1) idd = *open.begin();
    while (i < (int)pt.size() && pt[i].x == currx && pt[i].endpoint) {
      open.erase(pt[i].segid);
      ++i;
    }
    if (open.size() == 1 && contain > 1) segs[*open.begin()].candidate = true;
    if (open.size() == 1 && contain == 1 && stc == 0)
      segs[*open.begin()].isocand = true;
    if (open.size() == 0 && contain == 1 && stc == 0) segs[idd].iso = true;
    if (open.size() == 0) ++segcnt;
  }
  int maxc = 0;
  bool onlyiso = true;
  for (int i = 0; i < n; ++i) {
    maxc = max(maxc, segs[i].cnt);
    if (!segs[i].iso) onlyiso = false;
  }
  if (onlyiso)
    cout << segcnt - 1 << endl;
  else
    cout << segcnt + maxc << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> t;
  while (t--) solve();
  return 0;
}

```