#include <bits/stdc++.h>
using namespace std;
int N;
int cnt[202020];
int B[202020];
pair<int, int> A[202020];
vector<pair<int, int> > evt;
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int N;
    scanf("%d", &N);
    for (int i = 1; i <= N; i++) scanf("%d%d", &A[i].first, &A[i].second);
    for (int i = 1; i <= N; i++) {
      B[i] = -1;
      cnt[i] = 0;
    }
    sort(A + 1, A + N + 1);
    set<int> S;
    for (int i = 1; i <= N; i++) {
      evt.push_back({A[i].first, -i});
      evt.push_back({A[i].second, i});
    }
    sort(evt.begin(), evt.end());
    int ans = 0, mx = 0;
    int tmp = 0;
    for (pair<int, int> i : evt) {
      int id = abs(i.second);
      if (i.second < 0) {
        if (S.size() == 1 && tmp > 1) {
          B[id] = *(S.begin());
        } else if (S.size() == 0) {
          ans++;
          tmp = 0;
        }
        S.insert(id);
        tmp++;
      } else {
        S.erase(id);
      }
    }
    for (int i = 1; i <= N; i++)
      if (B[i] != -1) cnt[B[i]]++;
    for (int i = 1; i <= N; i++) {
      mx = max(mx, cnt[i]);
    }
    if (ans == N)
      printf("%d\n", N - 1);
    else
      printf("%d\n", mx + ans);
    evt.clear();
  }
  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 N;
int cnt[202020];
int B[202020];
pair<int, int> A[202020];
vector<pair<int, int> > evt;
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int N;
    scanf("%d", &N);
    for (int i = 1; i <= N; i++) scanf("%d%d", &A[i].first, &A[i].second);
    for (int i = 1; i <= N; i++) {
      B[i] = -1;
      cnt[i] = 0;
    }
    sort(A + 1, A + N + 1);
    set<int> S;
    for (int i = 1; i <= N; i++) {
      evt.push_back({A[i].first, -i});
      evt.push_back({A[i].second, i});
    }
    sort(evt.begin(), evt.end());
    int ans = 0, mx = 0;
    int tmp = 0;
    for (pair<int, int> i : evt) {
      int id = abs(i.second);
      if (i.second < 0) {
        if (S.size() == 1 && tmp > 1) {
          B[id] = *(S.begin());
        } else if (S.size() == 0) {
          ans++;
          tmp = 0;
        }
        S.insert(id);
        tmp++;
      } else {
        S.erase(id);
      }
    }
    for (int i = 1; i <= N; i++)
      if (B[i] != -1) cnt[B[i]]++;
    for (int i = 1; i <= N; i++) {
      mx = max(mx, cnt[i]);
    }
    if (ans == N)
      printf("%d\n", N - 1);
    else
      printf("%d\n", mx + ans);
    evt.clear();
  }
  return 0;
}

```