#include <bits/stdc++.h>
using namespace std;
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
void file() {}
void fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
}
const int INF = 2e9;
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int> > v1(n);
  vector<int> dif(n);
  map<int, int> ls;
  for (auto& it : v1) cin >> it.first >> it.second;
  sort(v1.begin(), v1.end());
  int l = INF, r = -INF, count = 0;
  for (int i = 0; i < n; i++) {
    if (v1[i].first > r) {
      if (r != -INF) ls[l] = 0;
      count++;
      l = v1[i].first, r = v1[i].second;
    } else
      r = max(r, v1[i].second);
  }
  ls[l] = 0;
  set<pair<int, int> > st;
  for (int i = 0; i < n; i++) {
    while (!st.empty() && st.begin()->first < v1[i].first) st.erase(st.begin());
    int j = i, cnt = 0;
    while (j < n && v1[j].first == v1[i].first) j++, cnt++;
    if (cnt > 1) {
      j = i;
      if ((int)(st.size()) == 1) {
        dif[st.begin()->second]++;
      }
      while (j < n && v1[j].first == v1[i].first) {
        st.insert({v1[j].second, j});
        j++;
      }
      i = j - 1;
      continue;
    }
    if ((int)(st.size()) == 1) {
      dif[st.begin()->second]++;
    }
    st.insert({v1[i].second, i});
  }
  for (int i = 0; i < n; i++)
    if (ls.count(v1[i].first)) ++ls[v1[i].first];
  for (int i = 0; i < n; i++)
    if (ls[v1[i].first] == 1) --dif[i];
  cout << count + *max_element(dif.begin(), dif.end()) << "\n";
}
int main() {
  file();
  fast();
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
void file() {}
void fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
}
const int INF = 2e9;
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int> > v1(n);
  vector<int> dif(n);
  map<int, int> ls;
  for (auto& it : v1) cin >> it.first >> it.second;
  sort(v1.begin(), v1.end());
  int l = INF, r = -INF, count = 0;
  for (int i = 0; i < n; i++) {
    if (v1[i].first > r) {
      if (r != -INF) ls[l] = 0;
      count++;
      l = v1[i].first, r = v1[i].second;
    } else
      r = max(r, v1[i].second);
  }
  ls[l] = 0;
  set<pair<int, int> > st;
  for (int i = 0; i < n; i++) {
    while (!st.empty() && st.begin()->first < v1[i].first) st.erase(st.begin());
    int j = i, cnt = 0;
    while (j < n && v1[j].first == v1[i].first) j++, cnt++;
    if (cnt > 1) {
      j = i;
      if ((int)(st.size()) == 1) {
        dif[st.begin()->second]++;
      }
      while (j < n && v1[j].first == v1[i].first) {
        st.insert({v1[j].second, j});
        j++;
      }
      i = j - 1;
      continue;
    }
    if ((int)(st.size()) == 1) {
      dif[st.begin()->second]++;
    }
    st.insert({v1[i].second, i});
  }
  for (int i = 0; i < n; i++)
    if (ls.count(v1[i].first)) ++ls[v1[i].first];
  for (int i = 0; i < n; i++)
    if (ls[v1[i].first] == 1) --dif[i];
  cout << count + *max_element(dif.begin(), dif.end()) << "\n";
}
int main() {
  file();
  fast();
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```