#include <bits/stdc++.h>
using namespace std;
enum endpoint_t {
  kLeft = 0,
  kRight = 1,
};
struct Endpoint {
  int coordinate;
  endpoint_t type;
  int segment_id;
};
inline bool cmpl(const Endpoint& lhs, const Endpoint& rhs) {
  return make_pair(lhs.coordinate, lhs.type) <
         make_pair(rhs.coordinate, rhs.type);
}
inline bool cmpr(const Endpoint& lhs, const Endpoint& rhs) {
  return make_pair(rhs.coordinate, rhs.type) <
         make_pair(lhs.coordinate, lhs.type);
}
int solve(vector<Endpoint>& endpoints, int n) {
  vector<int> cnt(n, 0);
  int nc = 0;
  set<int> active_segments;
  sort(endpoints.begin(), endpoints.end(), cmpl);
  for (int i = 0; i < (n << 1);) {
    int coordinate = endpoints[i].coordinate;
    int j = i;
    while (j < (n << 1) && coordinate == endpoints[j].coordinate) ++j;
    int lc = 0;
    for (int k = i; k < j; ++k) {
      if (endpoints[k].type == kLeft) {
        active_segments.insert(endpoints[k].segment_id);
        ++lc;
      }
    }
    if (active_segments.size() == lc && lc == 1)
      for (auto active_segment : active_segments) --cnt[active_segment];
    set<int> to_erase;
    int rc = 0;
    for (int k = i; k < j; ++k) {
      if (endpoints[k].type == kRight) {
        to_erase.insert(endpoints[k].segment_id);
        ++rc;
      }
    }
    if (active_segments.size() == rc && rc == 1)
      for (auto active_segment : active_segments) --cnt[active_segment];
    for (auto to_erase_i : to_erase) active_segments.erase(to_erase_i);
    if (active_segments.empty()) ++nc;
    if (active_segments.size() == 1)
      for (auto active_segment : active_segments) ++cnt[active_segment];
    i = j;
  }
  for (int i = 0; i < n; ++i)
    if (cnt[i] == -2) cnt[i] = -1;
  return nc + *max_element(cnt.begin(), cnt.end());
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t-- > 0) {
    int n;
    cin >> n;
    vector<Endpoint> endpoints(n << 1);
    for (int i = 0; i < n; ++i) {
      int l, r;
      cin >> l >> r;
      endpoints[i << 1] = {l, kLeft, i};
      endpoints[(i << 1) | 1] = {r, kRight, i};
    }
    cout << solve(endpoints, n) << "\n";
  }
  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;
enum endpoint_t {
  kLeft = 0,
  kRight = 1,
};
struct Endpoint {
  int coordinate;
  endpoint_t type;
  int segment_id;
};
inline bool cmpl(const Endpoint& lhs, const Endpoint& rhs) {
  return make_pair(lhs.coordinate, lhs.type) <
         make_pair(rhs.coordinate, rhs.type);
}
inline bool cmpr(const Endpoint& lhs, const Endpoint& rhs) {
  return make_pair(rhs.coordinate, rhs.type) <
         make_pair(lhs.coordinate, lhs.type);
}
int solve(vector<Endpoint>& endpoints, int n) {
  vector<int> cnt(n, 0);
  int nc = 0;
  set<int> active_segments;
  sort(endpoints.begin(), endpoints.end(), cmpl);
  for (int i = 0; i < (n << 1);) {
    int coordinate = endpoints[i].coordinate;
    int j = i;
    while (j < (n << 1) && coordinate == endpoints[j].coordinate) ++j;
    int lc = 0;
    for (int k = i; k < j; ++k) {
      if (endpoints[k].type == kLeft) {
        active_segments.insert(endpoints[k].segment_id);
        ++lc;
      }
    }
    if (active_segments.size() == lc && lc == 1)
      for (auto active_segment : active_segments) --cnt[active_segment];
    set<int> to_erase;
    int rc = 0;
    for (int k = i; k < j; ++k) {
      if (endpoints[k].type == kRight) {
        to_erase.insert(endpoints[k].segment_id);
        ++rc;
      }
    }
    if (active_segments.size() == rc && rc == 1)
      for (auto active_segment : active_segments) --cnt[active_segment];
    for (auto to_erase_i : to_erase) active_segments.erase(to_erase_i);
    if (active_segments.empty()) ++nc;
    if (active_segments.size() == 1)
      for (auto active_segment : active_segments) ++cnt[active_segment];
    i = j;
  }
  for (int i = 0; i < n; ++i)
    if (cnt[i] == -2) cnt[i] = -1;
  return nc + *max_element(cnt.begin(), cnt.end());
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t-- > 0) {
    int n;
    cin >> n;
    vector<Endpoint> endpoints(n << 1);
    for (int i = 0; i < n; ++i) {
      int l, r;
      cin >> l >> r;
      endpoints[i << 1] = {l, kLeft, i};
      endpoints[(i << 1) | 1] = {r, kRight, i};
    }
    cout << solve(endpoints, n) << "\n";
  }
  return 0;
}

```