#include <bits/stdc++.h>
using namespace std;
template <typename T>
struct CoordinateCompression {
  vector<T> o;
  void push(const T& x) { o.push_back(x); }
  int build() {
    sort(o.begin(), o.end());
    o.erase(unique(o.begin(), o.end()), o.end());
    return o.size();
  }
  int operator()(const T& x) {
    auto it = lower_bound(o.begin(), o.end(), x);
    return it == o.end() || x < *it ? -1 : it - o.begin();
  }
  T& operator[](int i) { return o[i]; }
};
int main() {
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  int tc;
  cin >> tc;
  while (tc--) {
    int n;
    cin >> n;
    vector<int> l(n), r(n);
    CoordinateCompression<int> cc;
    for (int i = 0; i < n; i++) {
      cin >> l[i] >> r[i];
      cc.push(l[i]);
      cc.push(r[i]);
    }
    int m = cc.build();
    vector<vector<int> > born(m), die(m);
    for (int i = 0; i < n; i++) {
      born[cc(l[i])].push_back(i);
      die[cc(r[i])].push_back(i);
    }
    vector<int> delta(n, -1);
    set<int> open;
    int global = 0;
    for (int i = 0; i < m; i++) {
      global += open.empty();
      bool support = open.size() + born[i].size() > 1;
      if (open.size() == 1 && born[i].size() > 0) {
        delta[*open.begin()]++;
      }
      for (int j : born[i]) {
        open.insert(j);
        if (support) delta[j]++;
      }
      for (int j : die[i]) open.erase(j);
    }
    int ans = global + *max_element(delta.begin(), delta.end());
    cout << ans << '\n';
  }
}

### 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;
template <typename T>
struct CoordinateCompression {
  vector<T> o;
  void push(const T& x) { o.push_back(x); }
  int build() {
    sort(o.begin(), o.end());
    o.erase(unique(o.begin(), o.end()), o.end());
    return o.size();
  }
  int operator()(const T& x) {
    auto it = lower_bound(o.begin(), o.end(), x);
    return it == o.end() || x < *it ? -1 : it - o.begin();
  }
  T& operator[](int i) { return o[i]; }
};
int main() {
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  int tc;
  cin >> tc;
  while (tc--) {
    int n;
    cin >> n;
    vector<int> l(n), r(n);
    CoordinateCompression<int> cc;
    for (int i = 0; i < n; i++) {
      cin >> l[i] >> r[i];
      cc.push(l[i]);
      cc.push(r[i]);
    }
    int m = cc.build();
    vector<vector<int> > born(m), die(m);
    for (int i = 0; i < n; i++) {
      born[cc(l[i])].push_back(i);
      die[cc(r[i])].push_back(i);
    }
    vector<int> delta(n, -1);
    set<int> open;
    int global = 0;
    for (int i = 0; i < m; i++) {
      global += open.empty();
      bool support = open.size() + born[i].size() > 1;
      if (open.size() == 1 && born[i].size() > 0) {
        delta[*open.begin()]++;
      }
      for (int j : born[i]) {
        open.insert(j);
        if (support) delta[j]++;
      }
      for (int j : die[i]) open.erase(j);
    }
    int ans = global + *max_element(delta.begin(), delta.end());
    cout << ans << '\n';
  }
}

```