#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
struct Event {
  long long x, t, i;
  Event() {}
  Event(long long x, long long t, long long i) : x(x), t(t), i(i) {}
};
bool cmpEvents(const Event &a, const Event &b) {
  return tie(a.x, a.t, a.i) < tie(b.x, b.t, b.i);
}
const long long MAX_N = 2e5 + 5;
long long n;
long long cnt[MAX_N];
pair<bool, bool> cover[MAX_N];
void solve() {
  cin >> n;
  vector<Event> sc;
  for (long long i = 1; i <= n; ++i) {
    cnt[i] = 0;
    cover[i] = make_pair(false, false);
    long long l, r;
    cin >> l >> r;
    l *= 2, r *= 2;
    sc.emplace_back(l, -1, i);
    sc.emplace_back(r, 1, i);
  }
  sort(sc.begin(), sc.end(), cmpEvents);
  for (long long i = 1; i < 2 * n; ++i) {
    sc.emplace_back((sc[i - 1].x + sc[i].x) / 2, 0, 0);
  }
  sort(sc.begin(), sc.end(), cmpEvents);
  set<long long> alive;
  long long start_ans = 2;
  for (long long i = 0; i < (long long)sc.size();) {
    long long j = i;
    while (j < (long long)sc.size() && sc[i].x == sc[j].x && sc[j].t == -1) {
      alive.emplace(sc[j].i);
      ++j;
    }
    while (i < j) {
      cover[sc[i].i].first = (long long)alive.size() > 1;
      ++i;
    }
    while (j < (long long)sc.size() && sc[i].x == sc[j].x && sc[j].t == 0) {
      if ((long long)alive.size() == 1) {
        cnt[*alive.begin()]++;
      }
      ++j;
    }
    if (alive.empty()) {
      start_ans++;
    }
    i = j;
    while (j < (long long)sc.size() && sc[i].x == sc[j].x && sc[j].t == 1) {
      cover[sc[j].i].second = (long long)alive.size() > 1;
      ++j;
    }
    while (i < j) {
      alive.erase(sc[i].i);
      ++i;
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= n; ++i) {
    ans = max(ans, start_ans + cnt[i] - !cover[i].first - !cover[i].second - 1);
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
struct Event {
  long long x, t, i;
  Event() {}
  Event(long long x, long long t, long long i) : x(x), t(t), i(i) {}
};
bool cmpEvents(const Event &a, const Event &b) {
  return tie(a.x, a.t, a.i) < tie(b.x, b.t, b.i);
}
const long long MAX_N = 2e5 + 5;
long long n;
long long cnt[MAX_N];
pair<bool, bool> cover[MAX_N];
void solve() {
  cin >> n;
  vector<Event> sc;
  for (long long i = 1; i <= n; ++i) {
    cnt[i] = 0;
    cover[i] = make_pair(false, false);
    long long l, r;
    cin >> l >> r;
    l *= 2, r *= 2;
    sc.emplace_back(l, -1, i);
    sc.emplace_back(r, 1, i);
  }
  sort(sc.begin(), sc.end(), cmpEvents);
  for (long long i = 1; i < 2 * n; ++i) {
    sc.emplace_back((sc[i - 1].x + sc[i].x) / 2, 0, 0);
  }
  sort(sc.begin(), sc.end(), cmpEvents);
  set<long long> alive;
  long long start_ans = 2;
  for (long long i = 0; i < (long long)sc.size();) {
    long long j = i;
    while (j < (long long)sc.size() && sc[i].x == sc[j].x && sc[j].t == -1) {
      alive.emplace(sc[j].i);
      ++j;
    }
    while (i < j) {
      cover[sc[i].i].first = (long long)alive.size() > 1;
      ++i;
    }
    while (j < (long long)sc.size() && sc[i].x == sc[j].x && sc[j].t == 0) {
      if ((long long)alive.size() == 1) {
        cnt[*alive.begin()]++;
      }
      ++j;
    }
    if (alive.empty()) {
      start_ans++;
    }
    i = j;
    while (j < (long long)sc.size() && sc[i].x == sc[j].x && sc[j].t == 1) {
      cover[sc[j].i].second = (long long)alive.size() > 1;
      ++j;
    }
    while (i < j) {
      alive.erase(sc[i].i);
      ++i;
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= n; ++i) {
    ans = max(ans, start_ans + cnt[i] - !cover[i].first - !cover[i].second - 1);
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

```