#include <bits/stdc++.h>
using namespace std;
int inf = 2e9;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
map<int, int> tab;
int getan(vector<pair<int, int> > a) {
  int cnt = 0;
  int l = -inf, r = -inf;
  sort(a.begin(), a.end());
  for (int i = 0; i < a.size(); i++) {
    if (a[i].first > r) {
      if (r != -inf) tab[l] = 0;
      l = a[i].first, r = a[i].second;
      cnt++;
    } else {
      r = max(r, a[i].second);
    }
  }
  tab[l] = 0;
  return cnt;
}
void fun(vector<pair<int, pair<int, int> > > a, vector<int> &ans) {
  set<int> an;
  for (int i = 0; i < a.size(); i++) {
    vector<int> o, e;
    int j = i - 1;
    while (j + 1 < a.size() && a[j + 1].first == a[i].first) {
      ++j;
      if (a[j].second.second == 1)
        o.push_back(a[j].second.first);
      else
        e.push_back(a[j].second.first);
    }
    if (an.size() == 1 && !o.empty()) {
      ans[*an.begin()]++;
    }
    for (int k = 0; k < o.size(); k++) {
      an.insert(o[k]);
    }
    for (int k = 0; k < e.size(); k++) {
      an.erase(e[k]);
    }
    i = j;
  }
}
void solve() {
  tab.clear();
  int k;
  cin >> k;
  vector<pair<int, int> > p(k);
  vector<pair<int, pair<int, int> > > f;
  vector<int> ans(k, 0);
  for (int i = 0; i < k; i++) {
    cin >> p[i].first >> p[i].second;
  }
  for (int i = 0; i < k; i++) {
    f.push_back(make_pair(p[i].first, make_pair(i, 1)));
    f.push_back(make_pair(p[i].second, make_pair(i, -1)));
  }
  sort(f.begin(), f.end());
  int cur = getan(p);
  fun(f, ans);
  for (int i = 0; i < k; i++)
    if (tab.count(p[i].first)) tab[p[i].first]++;
  for (int i = 0; i < k; i++)
    if (tab[p[i].first] == 1) ans[i]--;
  cout << cur + (*max_element(ans.begin(), ans.end())) << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) solve();
  return 0;
}

### 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;
int inf = 2e9;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
map<int, int> tab;
int getan(vector<pair<int, int> > a) {
  int cnt = 0;
  int l = -inf, r = -inf;
  sort(a.begin(), a.end());
  for (int i = 0; i < a.size(); i++) {
    if (a[i].first > r) {
      if (r != -inf) tab[l] = 0;
      l = a[i].first, r = a[i].second;
      cnt++;
    } else {
      r = max(r, a[i].second);
    }
  }
  tab[l] = 0;
  return cnt;
}
void fun(vector<pair<int, pair<int, int> > > a, vector<int> &ans) {
  set<int> an;
  for (int i = 0; i < a.size(); i++) {
    vector<int> o, e;
    int j = i - 1;
    while (j + 1 < a.size() && a[j + 1].first == a[i].first) {
      ++j;
      if (a[j].second.second == 1)
        o.push_back(a[j].second.first);
      else
        e.push_back(a[j].second.first);
    }
    if (an.size() == 1 && !o.empty()) {
      ans[*an.begin()]++;
    }
    for (int k = 0; k < o.size(); k++) {
      an.insert(o[k]);
    }
    for (int k = 0; k < e.size(); k++) {
      an.erase(e[k]);
    }
    i = j;
  }
}
void solve() {
  tab.clear();
  int k;
  cin >> k;
  vector<pair<int, int> > p(k);
  vector<pair<int, pair<int, int> > > f;
  vector<int> ans(k, 0);
  for (int i = 0; i < k; i++) {
    cin >> p[i].first >> p[i].second;
  }
  for (int i = 0; i < k; i++) {
    f.push_back(make_pair(p[i].first, make_pair(i, 1)));
    f.push_back(make_pair(p[i].second, make_pair(i, -1)));
  }
  sort(f.begin(), f.end());
  int cur = getan(p);
  fun(f, ans);
  for (int i = 0; i < k; i++)
    if (tab.count(p[i].first)) tab[p[i].first]++;
  for (int i = 0; i < k; i++)
    if (tab[p[i].first] == 1) ans[i]--;
  cout << cur + (*max_element(ans.begin(), ans.end())) << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) solve();
  return 0;
}

```