#include <bits/stdc++.h>
using namespace std;
vector<pair<int, pair<int, int>>> qry;
vector<pair<int, int>> p;
map<int, int> l;
vector<int> ans;
int union_set() {
  int s = p[0].first, e = p[0].second, ret = 1;
  l[s] = 0;
  for (int i = 1; i < p.size(); ++i) {
    if (p[i].first > e) l[p[i].first] = 0, ret++;
    e = max(e, p[i].second);
  }
  return ret;
}
void calc_union() {
  set<pair<int, int>> s;
  sort(qry.begin(), qry.end());
  for (int i = 0; i < qry.size() - 1; ++i) {
    int j = i;
    while (qry[j].second.first == 0 && qry[j].first == qry[i].first) {
      int idx = qry[j++].second.second;
      s.erase({p[idx].first, idx});
    }
    if (s.size() == 1 && qry[i].first == qry[j].first)
      ans[(*s.begin()).second]++;
    while (qry[j].first == qry[i].first)
      s.insert({qry[j].first, qry[j++].second.second});
    i = j - 1;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    l.clear();
    p.resize(n);
    ans.assign(n, 0);
    qry.clear();
    for (int i = 0; i < n; ++i) cin >> p[i].first >> p[i].second;
    sort(p.begin(), p.end());
    for (int i = 0; i < n; ++i) {
      qry.push_back({p[i].first, {1, i}});
      qry.push_back({p[i].second + 1, {0, i}});
    }
    qry.push_back({1e9 + 5, {5, -1}});
    int init = union_set(), nf = INT_MIN;
    calc_union();
    for (int i = 0; i < n; ++i)
      if (l.count(p[i].first)) l[p[i].first]++;
    for (int i = 0; i < n; ++i) {
      if (l[p[i].first] == 1) ans[i]--;
      nf = max(ans[i], nf);
    }
    cout << init + nf << '\n';
  }
}

### Prompt

Create a solution in CPP for 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;
vector<pair<int, pair<int, int>>> qry;
vector<pair<int, int>> p;
map<int, int> l;
vector<int> ans;
int union_set() {
  int s = p[0].first, e = p[0].second, ret = 1;
  l[s] = 0;
  for (int i = 1; i < p.size(); ++i) {
    if (p[i].first > e) l[p[i].first] = 0, ret++;
    e = max(e, p[i].second);
  }
  return ret;
}
void calc_union() {
  set<pair<int, int>> s;
  sort(qry.begin(), qry.end());
  for (int i = 0; i < qry.size() - 1; ++i) {
    int j = i;
    while (qry[j].second.first == 0 && qry[j].first == qry[i].first) {
      int idx = qry[j++].second.second;
      s.erase({p[idx].first, idx});
    }
    if (s.size() == 1 && qry[i].first == qry[j].first)
      ans[(*s.begin()).second]++;
    while (qry[j].first == qry[i].first)
      s.insert({qry[j].first, qry[j++].second.second});
    i = j - 1;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    l.clear();
    p.resize(n);
    ans.assign(n, 0);
    qry.clear();
    for (int i = 0; i < n; ++i) cin >> p[i].first >> p[i].second;
    sort(p.begin(), p.end());
    for (int i = 0; i < n; ++i) {
      qry.push_back({p[i].first, {1, i}});
      qry.push_back({p[i].second + 1, {0, i}});
    }
    qry.push_back({1e9 + 5, {5, -1}});
    int init = union_set(), nf = INT_MIN;
    calc_union();
    for (int i = 0; i < n; ++i)
      if (l.count(p[i].first)) l[p[i].first]++;
    for (int i = 0; i < n; ++i) {
      if (l[p[i].first] == 1) ans[i]--;
      nf = max(ans[i], nf);
    }
    cout << init + nf << '\n';
  }
}

```