#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
int treen, t, n, tot, a[1000000], res[1000000];
pair<int, int> seg[200000 + 10];
map<int, int> lisan;
void mian() {
  memset(a, 0, sizeof(a));
  lisan.clear();
  tot = 0;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> seg[i].first >> seg[i].second;
    lisan[seg[i].first] = lisan[seg[i].second] = 1;
  }
  for (map<int, int>::iterator ite = lisan.begin(); ite != lisan.end(); ite++) {
    ite->second = ++tot;
  }
  int MAX = -1;
  for (int i = 1; i <= n; ++i) {
    seg[i].first = lisan[seg[i].first] * 2;
    seg[i].second = lisan[seg[i].second] * 2;
    MAX = max(MAX, max(seg[i].first, seg[i].second));
  }
  MAX++;
  for (int i = 1; i <= MAX; ++i) res[i] = a[i] = 0;
  sort(seg + 1, seg + 1 + n);
  int OK = 0;
  for (int i = 2; i <= n; ++i) {
    if (seg[i - 1].second >= seg[i].first) {
      OK = 1;
      break;
    }
  }
  if (!OK) {
    cout << n - 1 << endl;
    return;
  }
  for (int i = 1; i <= n; ++i) {
    a[seg[i].first]++;
    a[seg[i].second + 1]--;
  }
  int res_ = 0;
  for (int i = 1; i <= MAX; ++i) a[i] += a[i - 1];
  for (int i = 1; i <= MAX; ++i) {
    if (a[i] == 0 && a[i - 1]) {
      res_++;
    }
    if (a[i] == 1 && a[i - 1] > 1 && a[i + 1] > 1) {
      res[i]++;
    }
  }
  int add = 0;
  for (int i = 1; i <= MAX; ++i) {
    res[i] += res[i - 1];
  }
  for (int i = 1; i <= n; ++i) {
    add = max(add, res[seg[i].second] - res[seg[i].first - 1]);
  }
  cout << res_ + add << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> t;
  while (t--) mian();
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
const int INF = 0x3f3f3f3f;
int treen, t, n, tot, a[1000000], res[1000000];
pair<int, int> seg[200000 + 10];
map<int, int> lisan;
void mian() {
  memset(a, 0, sizeof(a));
  lisan.clear();
  tot = 0;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> seg[i].first >> seg[i].second;
    lisan[seg[i].first] = lisan[seg[i].second] = 1;
  }
  for (map<int, int>::iterator ite = lisan.begin(); ite != lisan.end(); ite++) {
    ite->second = ++tot;
  }
  int MAX = -1;
  for (int i = 1; i <= n; ++i) {
    seg[i].first = lisan[seg[i].first] * 2;
    seg[i].second = lisan[seg[i].second] * 2;
    MAX = max(MAX, max(seg[i].first, seg[i].second));
  }
  MAX++;
  for (int i = 1; i <= MAX; ++i) res[i] = a[i] = 0;
  sort(seg + 1, seg + 1 + n);
  int OK = 0;
  for (int i = 2; i <= n; ++i) {
    if (seg[i - 1].second >= seg[i].first) {
      OK = 1;
      break;
    }
  }
  if (!OK) {
    cout << n - 1 << endl;
    return;
  }
  for (int i = 1; i <= n; ++i) {
    a[seg[i].first]++;
    a[seg[i].second + 1]--;
  }
  int res_ = 0;
  for (int i = 1; i <= MAX; ++i) a[i] += a[i - 1];
  for (int i = 1; i <= MAX; ++i) {
    if (a[i] == 0 && a[i - 1]) {
      res_++;
    }
    if (a[i] == 1 && a[i - 1] > 1 && a[i + 1] > 1) {
      res[i]++;
    }
  }
  int add = 0;
  for (int i = 1; i <= MAX; ++i) {
    res[i] += res[i - 1];
  }
  for (int i = 1; i <= n; ++i) {
    add = max(add, res[seg[i].second] - res[seg[i].first - 1]);
  }
  cout << res_ + add << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> t;
  while (t--) mian();
  return 0;
}

```