#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int INF = 1e9;
int main() {
  int t;
  cin >> t;
  vector<int> ans(t);
  vector<int> a;
  for (int i = 0; i < (int)t; i++) {
    int n;
    cin >> n;
    a.clear();
    int l[n], r[n];
    set<int> s;
    for (int j = 0; j < (int)n; j++) {
      cin >> l[j] >> r[j];
      if (s.find(l[j]) == s.end()) a.push_back(l[j]);
      s.insert(l[j]);
      if (s.find(r[j]) == s.end()) a.push_back(r[j]);
      s.insert(r[j]);
    }
    sort(a.begin(), a.end());
    int v = s.size();
    map<int, int> m;
    for (int j = 0; j < (int)v; j++) m[a[j]] = j * 2 + 1;
    vector<int> seg(2 * v + 1, 0);
    vector<int> one(2 * v + 1, 0);
    for (int j = 0; j < (int)n; j++) {
      seg[m[l[j]]]++;
      seg[m[r[j]] + 1]--;
    }
    for (int j = 1; j <= 2 * v; j++) seg[j] += seg[j - 1];
    int p = 1;
    for (int j = 1; j < 2 * v; j++) {
      if (seg[j] == 0) p++;
    }
    for (int j = 1; j <= 2 * v; j++) {
      one[j] += one[j - 1];
      if (j % 2 == 0 && seg[j] == 1) one[j]++;
    }
    int b = -INF;
    for (int j = 0; j < (int)n; j++) {
      int dif = one[m[r[j]]] - one[m[l[j]]];
      if (seg[m[r[j]] + 1] == 0 && seg[m[r[j]]] == 1 && seg[m[r[j]] - 1] == 1)
        dif--;
      if (seg[m[l[j]] - 1] == 0 && seg[m[l[j]]] == 1 && seg[m[l[j]] + 1] == 1)
        dif--;
      if (r[j] == l[j] && seg[m[r[j]]] == 1 && seg[m[r[j]] + 1] == 0 &&
          seg[m[l[j]] - 1] == 0)
        dif--;
      b = max(b, dif);
    }
    ans[i] = p + b;
  }
  for (auto k : ans) cout << k << endl;
  return 0;
}

### 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;
using ll = long long;
const int INF = 1e9;
int main() {
  int t;
  cin >> t;
  vector<int> ans(t);
  vector<int> a;
  for (int i = 0; i < (int)t; i++) {
    int n;
    cin >> n;
    a.clear();
    int l[n], r[n];
    set<int> s;
    for (int j = 0; j < (int)n; j++) {
      cin >> l[j] >> r[j];
      if (s.find(l[j]) == s.end()) a.push_back(l[j]);
      s.insert(l[j]);
      if (s.find(r[j]) == s.end()) a.push_back(r[j]);
      s.insert(r[j]);
    }
    sort(a.begin(), a.end());
    int v = s.size();
    map<int, int> m;
    for (int j = 0; j < (int)v; j++) m[a[j]] = j * 2 + 1;
    vector<int> seg(2 * v + 1, 0);
    vector<int> one(2 * v + 1, 0);
    for (int j = 0; j < (int)n; j++) {
      seg[m[l[j]]]++;
      seg[m[r[j]] + 1]--;
    }
    for (int j = 1; j <= 2 * v; j++) seg[j] += seg[j - 1];
    int p = 1;
    for (int j = 1; j < 2 * v; j++) {
      if (seg[j] == 0) p++;
    }
    for (int j = 1; j <= 2 * v; j++) {
      one[j] += one[j - 1];
      if (j % 2 == 0 && seg[j] == 1) one[j]++;
    }
    int b = -INF;
    for (int j = 0; j < (int)n; j++) {
      int dif = one[m[r[j]]] - one[m[l[j]]];
      if (seg[m[r[j]] + 1] == 0 && seg[m[r[j]]] == 1 && seg[m[r[j]] - 1] == 1)
        dif--;
      if (seg[m[l[j]] - 1] == 0 && seg[m[l[j]]] == 1 && seg[m[l[j]] + 1] == 1)
        dif--;
      if (r[j] == l[j] && seg[m[r[j]]] == 1 && seg[m[r[j]] + 1] == 0 &&
          seg[m[l[j]] - 1] == 0)
        dif--;
      b = max(b, dif);
    }
    ans[i] = p + b;
  }
  for (auto k : ans) cout << k << endl;
  return 0;
}

```