#include <bits/stdc++.h>
const int mod = 1000000007;
const int inf = 1000000009;
const long long INF = 1000000000000000009;
const long long big = 1000000000000000;
const long double eps = 0.000000000000000000001;
using namespace std;
vector<pair<pair<int, int>, int> > W;
vector<int> M;
pair<int, int> T[200005];
vector<pair<int, int> > K;
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int t;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    int n;
    cin >> n;
    for (int j = 1; j <= n; j++) {
      cin >> T[j].first >> T[j].second;
      W.push_back(make_pair(make_pair(T[j].first, 0), j));
      W.push_back(make_pair(make_pair(T[j].second, 1), j));
    }
    sort(W.begin(), W.end());
    int wynik = 0, maks = -inf, ile = 0;
    for (auto v : W) {
      if (v.first.second == 0) {
        ile++;
        if (v.first.first > maks) wynik++;
        for (auto x : K) {
          if (x.second >= v.first.first) M.push_back(x.first);
        }
        K.clear();
        maks = max(T[v.second].second, maks);
      } else {
        if (ile == 2) K.push_back(make_pair(v.first.first, maks));
        ile--;
      }
    }
    sort(M.begin(), M.end());
    M.resize(unique(M.begin(), M.end()) - M.begin());
    int aktual = 0;
    for (int j = 1; j <= n; j++) {
      int p = lower_bound(M.begin(), M.end(), T[j].first) - M.begin();
      int k = upper_bound(M.begin(), M.end(), T[j].second - 1) - M.begin();
      aktual = max(aktual, k - p);
    }
    wynik += aktual;
    wynik = min(wynik, n - 1);
    cout << wynik << "\n";
    W.clear();
    M.clear();
    K.clear();
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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>
const int mod = 1000000007;
const int inf = 1000000009;
const long long INF = 1000000000000000009;
const long long big = 1000000000000000;
const long double eps = 0.000000000000000000001;
using namespace std;
vector<pair<pair<int, int>, int> > W;
vector<int> M;
pair<int, int> T[200005];
vector<pair<int, int> > K;
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int t;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    int n;
    cin >> n;
    for (int j = 1; j <= n; j++) {
      cin >> T[j].first >> T[j].second;
      W.push_back(make_pair(make_pair(T[j].first, 0), j));
      W.push_back(make_pair(make_pair(T[j].second, 1), j));
    }
    sort(W.begin(), W.end());
    int wynik = 0, maks = -inf, ile = 0;
    for (auto v : W) {
      if (v.first.second == 0) {
        ile++;
        if (v.first.first > maks) wynik++;
        for (auto x : K) {
          if (x.second >= v.first.first) M.push_back(x.first);
        }
        K.clear();
        maks = max(T[v.second].second, maks);
      } else {
        if (ile == 2) K.push_back(make_pair(v.first.first, maks));
        ile--;
      }
    }
    sort(M.begin(), M.end());
    M.resize(unique(M.begin(), M.end()) - M.begin());
    int aktual = 0;
    for (int j = 1; j <= n; j++) {
      int p = lower_bound(M.begin(), M.end(), T[j].first) - M.begin();
      int k = upper_bound(M.begin(), M.end(), T[j].second - 1) - M.begin();
      aktual = max(aktual, k - p);
    }
    wynik += aktual;
    wynik = min(wynik, n - 1);
    cout << wynik << "\n";
    W.clear();
    M.clear();
    K.clear();
  }
  return 0;
}

```