#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  ios::sync_with_stdio(0), cin.tie(0);
  long long T;
  cin >> T;
  while (T--) {
    long long n;
    cin >> n;
    pair<long long, long long> v[n];
    map<long long, long long> m, pf, k;
    for (long long i = 0; i < n; i++) {
      cin >> v[i].first >> v[i].second;
      v[i].first *= 2, v[i].second *= 2;
      m[v[i].first]++, m[v[i].second + 1]--;
      k[v[i].second]++;
    }
    long long p = 0, l = 0, ans = 0;
    for (auto i : m) {
      m[i.first] += p;
      p = m[i.first];
      if (m[i.first] == 0) l++;
    }
    p = 0;
    for (auto i : m) {
      if (i.second == 1) p++;
      pf[i.first] = p;
    }
    for (auto i : v) {
      auto it = --pf.find(i.second + 1);
      long long x = pf[(*it).first] - pf[i.first];
      if ((*it).first == i.first and m[i.first] == 1)
        x = -1;
      else if (m[(*it).first] == 1 and m[i.second + 1] == 0)
        x--;
      ans = max(ans, l + x);
    }
    cout << ans << "\n";
  }
}

### Prompt

Generate 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;
int32_t main() {
  ios::sync_with_stdio(0), cin.tie(0);
  long long T;
  cin >> T;
  while (T--) {
    long long n;
    cin >> n;
    pair<long long, long long> v[n];
    map<long long, long long> m, pf, k;
    for (long long i = 0; i < n; i++) {
      cin >> v[i].first >> v[i].second;
      v[i].first *= 2, v[i].second *= 2;
      m[v[i].first]++, m[v[i].second + 1]--;
      k[v[i].second]++;
    }
    long long p = 0, l = 0, ans = 0;
    for (auto i : m) {
      m[i.first] += p;
      p = m[i.first];
      if (m[i.first] == 0) l++;
    }
    p = 0;
    for (auto i : m) {
      if (i.second == 1) p++;
      pf[i.first] = p;
    }
    for (auto i : v) {
      auto it = --pf.find(i.second + 1);
      long long x = pf[(*it).first] - pf[i.first];
      if ((*it).first == i.first and m[i.first] == 1)
        x = -1;
      else if (m[(*it).first] == 1 and m[i.second + 1] == 0)
        x--;
      ans = max(ans, l + x);
    }
    cout << ans << "\n";
  }
}

```