#include <bits/stdc++.h>
using namespace std;
int q;
int main() {
  cin >> q;
  while (q--) {
    map<int, int> mp;
    set<int> se;
    vector<pair<int, int> > v;
    vector<pair<int, pair<int, int> > > v1;
    int x, y, i, n, ans[+200500]{0}, r, mx, kl;
    mp.clear();
    se.clear();
    v.clear();
    v1.clear();
    cin >> n;
    for (i = 1; i <= n; i++) cin >> x >> y, v.push_back({x, y});
    sort(v.begin(), v.end());
    for (i = 0; i < n; i++)
      v1.push_back({v[i].first, {i, 1}}), v1.push_back({v[i].second, {i, -1}});
    sort(v1.begin(), v1.end());
    i = 0;
    while (i < v1.size()) {
      vector<int> vo, vc;
      vo.clear();
      vc.clear();
      int j = i;
      while (j < v1.size() && v1[j].first == v1[i].first) {
        if (v1[j].second.second == 1)
          vo.push_back(v1[j].second.first);
        else
          vc.push_back(v1[j].second.first);
        j++;
      }
      if (se.size() == 1 && vo.size()) ans[*se.begin()]++;
      for (auto it : vo) se.insert(it);
      for (auto it : vc) se.erase(it);
      i = j;
    }
    kl = 0;
    i = 0;
    int l = -2e9;
    r = -2e9;
    while (i < v.size()) {
      if (v[i].first > r) {
        if (r != -2e9) mp[l] = 0;
        kl++;
        l = v[i].first;
        r = v[i].second;
      } else
        r = max(r, v[i].second);
      i++;
    }
    mp[l] = 0;
    mx = -2e9;
    for (i = 0; i < v.size(); i++)
      if (mp.count(v[i].first)) mp[v[i].first]++;
    for (i = 0; i < v.size(); i++)
      if (mp[v[i].first] == 1) ans[i]--;
    for (i = 0; i < v.size(); i++) mx = max(mx, ans[i]);
    cout << kl + mx << endl;
  }
}

### Prompt

Your task is to create a Cpp solution 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;
int q;
int main() {
  cin >> q;
  while (q--) {
    map<int, int> mp;
    set<int> se;
    vector<pair<int, int> > v;
    vector<pair<int, pair<int, int> > > v1;
    int x, y, i, n, ans[+200500]{0}, r, mx, kl;
    mp.clear();
    se.clear();
    v.clear();
    v1.clear();
    cin >> n;
    for (i = 1; i <= n; i++) cin >> x >> y, v.push_back({x, y});
    sort(v.begin(), v.end());
    for (i = 0; i < n; i++)
      v1.push_back({v[i].first, {i, 1}}), v1.push_back({v[i].second, {i, -1}});
    sort(v1.begin(), v1.end());
    i = 0;
    while (i < v1.size()) {
      vector<int> vo, vc;
      vo.clear();
      vc.clear();
      int j = i;
      while (j < v1.size() && v1[j].first == v1[i].first) {
        if (v1[j].second.second == 1)
          vo.push_back(v1[j].second.first);
        else
          vc.push_back(v1[j].second.first);
        j++;
      }
      if (se.size() == 1 && vo.size()) ans[*se.begin()]++;
      for (auto it : vo) se.insert(it);
      for (auto it : vc) se.erase(it);
      i = j;
    }
    kl = 0;
    i = 0;
    int l = -2e9;
    r = -2e9;
    while (i < v.size()) {
      if (v[i].first > r) {
        if (r != -2e9) mp[l] = 0;
        kl++;
        l = v[i].first;
        r = v[i].second;
      } else
        r = max(r, v[i].second);
      i++;
    }
    mp[l] = 0;
    mx = -2e9;
    for (i = 0; i < v.size(); i++)
      if (mp.count(v[i].first)) mp[v[i].first]++;
    for (i = 0; i < v.size(); i++)
      if (mp[v[i].first] == 1) ans[i]--;
    for (i = 0; i < v.size(); i++) mx = max(mx, ans[i]);
    cout << kl + mx << endl;
  }
}

```