#include <bits/stdc++.h>
using namespace std;
int n;
pair<int, int> A[200003];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    if (n == 1) {
      printf("0\n");
      continue;
    }
    vector<pair<int, int> > upd;
    vector<pair<pair<int, int>, int> > vec;
    for (int i = 0; i < n; i++) {
      scanf("%d %d", &A[i].first, &A[i].second);
      A[i].first += A[i].first;
      A[i].second += A[i].second;
      upd.push_back({A[i].first, 1});
      upd.push_back({A[i].second + 1, -1});
    }
    sort(upd.begin(), upd.end());
    int kir = -2e9 - 5, kan, curr = 0;
    for (auto x : upd) {
      if (kir != -2e9 - 5 && x.first != kir)
        vec.push_back({{kir, x.first - 1}, curr});
      kir = x.first;
      curr += x.second;
    }
    int mx = 0, init = 1;
    for (auto x : vec)
      if (x.second == 0) init++;
    mx = init;
    vector<pair<int, int> > one;
    for (int i = 1; i < vec.size() - 1; i++) {
      if (vec[i].second == 1) {
        if ((vec[i + 1].second && vec[i - 1].second))
          one.push_back(vec[i].first);
      }
    }
    if (!one.size()) {
      printf("%d\n", min(init, n - 1));
      continue;
    }
    for (int i = 0; i < n; i++) {
      int X = A[i].first, Y = A[i].second;
      kir = 0, kan = one.size() - 1;
      while (kir < kan) {
        int mid = (kir + kan) / 2;
        if (one[mid].first < X)
          kir = mid + 1;
        else
          kan = mid;
      }
      int idx1 = kir;
      kir = 0, kan = one.size() - 1;
      while (kir < kan) {
        int mid = (kir + kan + 1) / 2;
        if (one[mid].second > Y)
          kan = mid - 1;
        else
          kir = mid;
      }
      int idx2 = kir;
      if (X <= one[idx1].first && one[idx1].second <= Y) {
        mx = max(mx, init + idx2 - idx1 + 1);
      }
    }
    printf("%d\n", min(mx, n - 1));
  }
}

### 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;
int n;
pair<int, int> A[200003];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    if (n == 1) {
      printf("0\n");
      continue;
    }
    vector<pair<int, int> > upd;
    vector<pair<pair<int, int>, int> > vec;
    for (int i = 0; i < n; i++) {
      scanf("%d %d", &A[i].first, &A[i].second);
      A[i].first += A[i].first;
      A[i].second += A[i].second;
      upd.push_back({A[i].first, 1});
      upd.push_back({A[i].second + 1, -1});
    }
    sort(upd.begin(), upd.end());
    int kir = -2e9 - 5, kan, curr = 0;
    for (auto x : upd) {
      if (kir != -2e9 - 5 && x.first != kir)
        vec.push_back({{kir, x.first - 1}, curr});
      kir = x.first;
      curr += x.second;
    }
    int mx = 0, init = 1;
    for (auto x : vec)
      if (x.second == 0) init++;
    mx = init;
    vector<pair<int, int> > one;
    for (int i = 1; i < vec.size() - 1; i++) {
      if (vec[i].second == 1) {
        if ((vec[i + 1].second && vec[i - 1].second))
          one.push_back(vec[i].first);
      }
    }
    if (!one.size()) {
      printf("%d\n", min(init, n - 1));
      continue;
    }
    for (int i = 0; i < n; i++) {
      int X = A[i].first, Y = A[i].second;
      kir = 0, kan = one.size() - 1;
      while (kir < kan) {
        int mid = (kir + kan) / 2;
        if (one[mid].first < X)
          kir = mid + 1;
        else
          kan = mid;
      }
      int idx1 = kir;
      kir = 0, kan = one.size() - 1;
      while (kir < kan) {
        int mid = (kir + kan + 1) / 2;
        if (one[mid].second > Y)
          kan = mid - 1;
        else
          kir = mid;
      }
      int idx2 = kir;
      if (X <= one[idx1].first && one[idx1].second <= Y) {
        mx = max(mx, init + idx2 - idx1 + 1);
      }
    }
    printf("%d\n", min(mx, n - 1));
  }
}

```