#include <bits/stdc++.h>
using namespace std;
int T;
int N;
signed long long L[202020], R[202020];
vector<int> ev[402020];
int always[201010], add[201010];
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  cin >> T;
  while (T--) {
    cin >> N;
    vector<signed long long> V;
    V.push_back(-(1LL << 60));
    for (i = 0; i < (N); i++) {
      cin >> L[i] >> R[i];
      L[i] *= 10000000;
      R[i] *= 10000000;
      R[i] += 1000000;
      V.push_back(L[i]);
      V.push_back(R[i]);
    }
    sort((V.begin()), (V.end()));
    for (i = 0; i < (N); i++) {
      always[i] = add[i] = 0;
      L[i] = lower_bound((V.begin()), (V.end()), L[i]) - V.begin();
      R[i] = lower_bound((V.begin()), (V.end()), R[i]) - V.begin();
      ev[L[i]].push_back(i);
      ev[R[i]].push_back(i);
    }
    set<int> S;
    int al = 0;
    for (i = 0; i < (V.size()); i++) {
      sort((ev[i].begin()), (ev[i].end()));
      for (auto& e : ev[i]) {
        if (i == L[e]) {
          if (S.empty()) {
            always[e]++;
            al++;
          } else if (S.size() == 1) {
            add[*S.begin()]++;
          }
          S.insert(e);
        } else {
          S.erase(e);
        }
      }
      ev[i].clear();
    }
    int ma = 0;
    for (i = 0; i < (N); i++) {
      ma = max(ma, al - always[i] + add[i]);
    }
    cout << ma << endl;
  }
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  cout.tie(0);
  solve();
  return 0;
}

### Prompt

Please formulate 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 T;
int N;
signed long long L[202020], R[202020];
vector<int> ev[402020];
int always[201010], add[201010];
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  cin >> T;
  while (T--) {
    cin >> N;
    vector<signed long long> V;
    V.push_back(-(1LL << 60));
    for (i = 0; i < (N); i++) {
      cin >> L[i] >> R[i];
      L[i] *= 10000000;
      R[i] *= 10000000;
      R[i] += 1000000;
      V.push_back(L[i]);
      V.push_back(R[i]);
    }
    sort((V.begin()), (V.end()));
    for (i = 0; i < (N); i++) {
      always[i] = add[i] = 0;
      L[i] = lower_bound((V.begin()), (V.end()), L[i]) - V.begin();
      R[i] = lower_bound((V.begin()), (V.end()), R[i]) - V.begin();
      ev[L[i]].push_back(i);
      ev[R[i]].push_back(i);
    }
    set<int> S;
    int al = 0;
    for (i = 0; i < (V.size()); i++) {
      sort((ev[i].begin()), (ev[i].end()));
      for (auto& e : ev[i]) {
        if (i == L[e]) {
          if (S.empty()) {
            always[e]++;
            al++;
          } else if (S.size() == 1) {
            add[*S.begin()]++;
          }
          S.insert(e);
        } else {
          S.erase(e);
        }
      }
      ev[i].clear();
    }
    int ma = 0;
    for (i = 0; i < (N); i++) {
      ma = max(ma, al - always[i] + add[i]);
    }
    cout << ma << endl;
  }
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  cout.tie(0);
  solve();
  return 0;
}

```