#include <bits/stdc++.h>
using namespace std;
void _init() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(10);
  cout << fixed;
}
int main() {
  _init();
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> d;
    vector<pair<int, int>> seg(n);
    for (auto &i : seg) {
      cin >> i.first >> i.second;
      d.push_back(i.first);
      d.push_back(i.second);
    }
    sort((d).begin(), (d).end());
    d.erase(unique((d).begin(), (d).end()), d.end());
    int mx = 0;
    for (auto &i : seg) {
      i.first = lower_bound((d).begin(), (d).end(), i.first) - d.begin() + 1;
      i.second = lower_bound((d).begin(), (d).end(), i.second) - d.begin() + 1;
      i.first *= 2;
      i.second *= 2;
      mx = max(mx, i.second);
    }
    mx += 2;
    vector<vector<int>> inc(mx), dec(mx);
    for (int i = 0; i < n; ++i) {
      inc[seg[i].first].push_back(i);
      dec[seg[i].second].push_back(i);
    }
    vector<pair<int, int>> cnt(mx);
    set<int> s;
    for (int i = 0; i < mx; ++i) {
      for (auto &j : inc[i]) {
        s.insert(j);
      }
      if ((int)(s).size() == 0)
        ;
      else if ((int)(s).size() == 1)
        cnt[i] = {1, *s.begin()};
      else
        cnt[i] = {2, -1};
      for (auto &j : dec[i]) {
        s.erase(j);
      }
    }
    int ori = 0;
    for (int i = 1; i < mx; ++i) {
      if (cnt[i - 1].first == 0 && cnt[i].first) ++ori;
    }
    if (ori == n) {
      cout << n - 1 << endl;
      continue;
    }
    vector<int> ans(n);
    for (int i = 0; i < mx;) {
      if (cnt[i].first == 1) {
        int j = i;
        while (cnt[j + 1].first == 1) ++j;
        if (cnt[i - 1].first == 2 && cnt[j + 1].first == 2)
          ans[cnt[i].second]++;
        i = j + 1;
      } else {
        ++i;
      }
    }
    cout << ori + *max_element((ans).begin(), (ans).end()) << endl;
  }
  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;
void _init() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(10);
  cout << fixed;
}
int main() {
  _init();
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> d;
    vector<pair<int, int>> seg(n);
    for (auto &i : seg) {
      cin >> i.first >> i.second;
      d.push_back(i.first);
      d.push_back(i.second);
    }
    sort((d).begin(), (d).end());
    d.erase(unique((d).begin(), (d).end()), d.end());
    int mx = 0;
    for (auto &i : seg) {
      i.first = lower_bound((d).begin(), (d).end(), i.first) - d.begin() + 1;
      i.second = lower_bound((d).begin(), (d).end(), i.second) - d.begin() + 1;
      i.first *= 2;
      i.second *= 2;
      mx = max(mx, i.second);
    }
    mx += 2;
    vector<vector<int>> inc(mx), dec(mx);
    for (int i = 0; i < n; ++i) {
      inc[seg[i].first].push_back(i);
      dec[seg[i].second].push_back(i);
    }
    vector<pair<int, int>> cnt(mx);
    set<int> s;
    for (int i = 0; i < mx; ++i) {
      for (auto &j : inc[i]) {
        s.insert(j);
      }
      if ((int)(s).size() == 0)
        ;
      else if ((int)(s).size() == 1)
        cnt[i] = {1, *s.begin()};
      else
        cnt[i] = {2, -1};
      for (auto &j : dec[i]) {
        s.erase(j);
      }
    }
    int ori = 0;
    for (int i = 1; i < mx; ++i) {
      if (cnt[i - 1].first == 0 && cnt[i].first) ++ori;
    }
    if (ori == n) {
      cout << n - 1 << endl;
      continue;
    }
    vector<int> ans(n);
    for (int i = 0; i < mx;) {
      if (cnt[i].first == 1) {
        int j = i;
        while (cnt[j + 1].first == 1) ++j;
        if (cnt[i - 1].first == 2 && cnt[j + 1].first == 2)
          ans[cnt[i].second]++;
        i = j + 1;
      } else {
        ++i;
      }
    }
    cout << ori + *max_element((ans).begin(), (ans).end()) << endl;
  }
  return 0;
}

```