#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n;
  cin >> n;
  vector<tuple<int, int, int>> v;
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    v.emplace_back(l, 1, i);
    v.emplace_back(r + 1, -1, i);
  }
  sort(v.begin(), v.end());
  set<int> s;
  int seg = 0, inf = 1000000001, m0 = inf;
  map<int, int> m;
  bool dbl = false;
  for (auto t : v) {
    int x = get<0>(t), d = get<1>(t), i = get<2>(t);
    if (d == 1) {
      if (s.empty()) {
        seg++;
      } else {
        dbl = true;
      }
      if (s.size() == 1 && *s.begin() == m0) m[m0]++;
      s.insert(i);
    } else {
      s.erase(i);
      if (s.size() == 1) m0 = *s.begin();
      if (s.empty()) m0 = inf;
    }
  }
  int mmax = 0;
  for (auto x : m) {
    mmax = max(mmax, x.second);
  }
  if (!dbl) seg--;
  cout << seg + mmax << endl;
}
int main() {
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) solve();
}

### 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;
void solve() {
  int n;
  cin >> n;
  vector<tuple<int, int, int>> v;
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    v.emplace_back(l, 1, i);
    v.emplace_back(r + 1, -1, i);
  }
  sort(v.begin(), v.end());
  set<int> s;
  int seg = 0, inf = 1000000001, m0 = inf;
  map<int, int> m;
  bool dbl = false;
  for (auto t : v) {
    int x = get<0>(t), d = get<1>(t), i = get<2>(t);
    if (d == 1) {
      if (s.empty()) {
        seg++;
      } else {
        dbl = true;
      }
      if (s.size() == 1 && *s.begin() == m0) m[m0]++;
      s.insert(i);
    } else {
      s.erase(i);
      if (s.size() == 1) m0 = *s.begin();
      if (s.empty()) m0 = inf;
    }
  }
  int mmax = 0;
  for (auto x : m) {
    mmax = max(mmax, x.second);
  }
  if (!dbl) seg--;
  cout << seg + mmax << endl;
}
int main() {
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) solve();
}

```