#include <bits/stdc++.h>
template <class T>
std::map<T, int> compress(std::vector<T>& v) {
  std::sort(v.begin(), v.end());
  v.erase(std::unique(v.begin(), v.end()), v.end());
  std::map<T, int> rev;
  for (int i = 0; i < (int)v.size(); ++i) {
    rev[v[i]] = i;
  }
  return rev;
}
constexpr int INF = 1 << 30;
void solve() {
  int n;
  std::cin >> n;
  std::vector<int> ls(n), rs(n), xs{-INF, INF};
  for (int i = 0; i < n; ++i) {
    std::cin >> ls[i] >> rs[i];
    xs.push_back(ls[i]);
    xs.push_back(rs[i]);
  }
  auto rev = compress(xs);
  for (auto& l : ls) l = rev[l] * 2;
  for (auto& r : rs) r = rev[r] * 2;
  int m = xs.size() * 2;
  std::vector<int> cnt(m, 0);
  for (int i = 0; i < n; ++i) {
    ++cnt[ls[i]];
    --cnt[rs[i] + 1];
  }
  for (int i = 1; i < m; ++i) cnt[i] += cnt[i - 1];
  int total = 0;
  for (int i = 0; i < m - 1; ++i) {
    if (cnt[i] == 0 && cnt[i + 1] > 0) ++total;
  }
  std::vector<int> inccnt(m - 1, 0);
  for (int i = 0; i < m - 1; ++i) {
    if (i > 0) inccnt[i] = inccnt[i - 1];
    if (cnt[i] == 1 && cnt[i + 1] > 1) ++inccnt[i];
  }
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    int inc = total;
    if (ls[i] < rs[i]) inc += inccnt[rs[i] - 1] - inccnt[ls[i] - 1];
    if (cnt[ls[i] - 1] == 0 && cnt[ls[i]] == 1) --inc;
    if (cnt[rs[i]] == 1 && cnt[rs[i] + 1] > 0) ++inc;
    ans = std::max(ans, inc);
  }
  std::cout << ans << "\n";
}
int main() {
  std::cin.tie(nullptr);
  std::ios::sync_with_stdio(false);
  int q;
  std::cin >> q;
  while (q--) solve();
  return 0;
}

### Prompt

Your challenge is to write 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>
template <class T>
std::map<T, int> compress(std::vector<T>& v) {
  std::sort(v.begin(), v.end());
  v.erase(std::unique(v.begin(), v.end()), v.end());
  std::map<T, int> rev;
  for (int i = 0; i < (int)v.size(); ++i) {
    rev[v[i]] = i;
  }
  return rev;
}
constexpr int INF = 1 << 30;
void solve() {
  int n;
  std::cin >> n;
  std::vector<int> ls(n), rs(n), xs{-INF, INF};
  for (int i = 0; i < n; ++i) {
    std::cin >> ls[i] >> rs[i];
    xs.push_back(ls[i]);
    xs.push_back(rs[i]);
  }
  auto rev = compress(xs);
  for (auto& l : ls) l = rev[l] * 2;
  for (auto& r : rs) r = rev[r] * 2;
  int m = xs.size() * 2;
  std::vector<int> cnt(m, 0);
  for (int i = 0; i < n; ++i) {
    ++cnt[ls[i]];
    --cnt[rs[i] + 1];
  }
  for (int i = 1; i < m; ++i) cnt[i] += cnt[i - 1];
  int total = 0;
  for (int i = 0; i < m - 1; ++i) {
    if (cnt[i] == 0 && cnt[i + 1] > 0) ++total;
  }
  std::vector<int> inccnt(m - 1, 0);
  for (int i = 0; i < m - 1; ++i) {
    if (i > 0) inccnt[i] = inccnt[i - 1];
    if (cnt[i] == 1 && cnt[i + 1] > 1) ++inccnt[i];
  }
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    int inc = total;
    if (ls[i] < rs[i]) inc += inccnt[rs[i] - 1] - inccnt[ls[i] - 1];
    if (cnt[ls[i] - 1] == 0 && cnt[ls[i]] == 1) --inc;
    if (cnt[rs[i]] == 1 && cnt[rs[i] + 1] > 0) ++inc;
    ans = std::max(ans, inc);
  }
  std::cout << ans << "\n";
}
int main() {
  std::cin.tie(nullptr);
  std::ios::sync_with_stdio(false);
  int q;
  std::cin >> q;
  while (q--) solve();
  return 0;
}

```