#include <bits/stdc++.h>
using namespace std;
struct Seg {
  int l;
  int r;
  bool operator<(const Seg& s) const { return l < s.l; }
} seg[200005];
struct Event {
  int tm;
  int i;
  int add;
  bool operator<(const Event& e) const {
    if (tm != e.tm) return tm < e.tm;
    return add < e.add;
  }
} event[200005 * 2];
int t, n;
int ans[200005];
map<int, int> x;
set<int> l;
vector<int> opens;
vector<int> closes;
int main() {
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 0; i < n; i++) {
      ans[i] = 0;
      cin >> seg[i].l >> seg[i].r;
    }
    sort(seg, seg + n);
    for (int i = 0; i < n; i++) {
      event[i * 2] = {seg[i].l, i, 1};
      event[i * 2 + 1] = {seg[i].r + 1, i, 0};
    }
    sort(event, event + 2 * n);
    l.clear();
    x.clear();
    for (int i = 0; i < 2 * n;) {
      opens.clear(), closes.clear();
      int j = i;
      while (j < 2 * n && event[j].tm == event[i].tm) {
        if (event[j].add)
          opens.push_back(event[j].i);
        else
          closes.push_back(event[j].i);
        j++;
      }
      i = j;
      for (int k : closes) l.erase(k);
      if (l.size() == 1 && !opens.empty()) {
        ans[*l.begin()]++;
      }
      for (int k : opens) l.insert(k);
    }
    int l = INT_MIN, r = INT_MIN;
    int num = 0;
    for (int i = 0; i < n; i++) {
      if (seg[i].l > r) {
        num++;
        x[seg[i].l] = 0;
        l = seg[i].l, r = seg[i].r;
      } else
        r = max(r, seg[i].r);
    }
    for (int i = 0; i < n; i++)
      if (x.find(seg[i].l) != x.end()) x[seg[i].l]++;
    for (int i = 0; i < n; i++) ans[i] -= (x[seg[i].l] == 1);
    cout << num + *max_element(ans, ans + n) << endl;
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
struct Seg {
  int l;
  int r;
  bool operator<(const Seg& s) const { return l < s.l; }
} seg[200005];
struct Event {
  int tm;
  int i;
  int add;
  bool operator<(const Event& e) const {
    if (tm != e.tm) return tm < e.tm;
    return add < e.add;
  }
} event[200005 * 2];
int t, n;
int ans[200005];
map<int, int> x;
set<int> l;
vector<int> opens;
vector<int> closes;
int main() {
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 0; i < n; i++) {
      ans[i] = 0;
      cin >> seg[i].l >> seg[i].r;
    }
    sort(seg, seg + n);
    for (int i = 0; i < n; i++) {
      event[i * 2] = {seg[i].l, i, 1};
      event[i * 2 + 1] = {seg[i].r + 1, i, 0};
    }
    sort(event, event + 2 * n);
    l.clear();
    x.clear();
    for (int i = 0; i < 2 * n;) {
      opens.clear(), closes.clear();
      int j = i;
      while (j < 2 * n && event[j].tm == event[i].tm) {
        if (event[j].add)
          opens.push_back(event[j].i);
        else
          closes.push_back(event[j].i);
        j++;
      }
      i = j;
      for (int k : closes) l.erase(k);
      if (l.size() == 1 && !opens.empty()) {
        ans[*l.begin()]++;
      }
      for (int k : opens) l.insert(k);
    }
    int l = INT_MIN, r = INT_MIN;
    int num = 0;
    for (int i = 0; i < n; i++) {
      if (seg[i].l > r) {
        num++;
        x[seg[i].l] = 0;
        l = seg[i].l, r = seg[i].r;
      } else
        r = max(r, seg[i].r);
    }
    for (int i = 0; i < n; i++)
      if (x.find(seg[i].l) != x.end()) x[seg[i].l]++;
    for (int i = 0; i < n; i++) ans[i] -= (x[seg[i].l] == 1);
    cout << num + *max_element(ans, ans + n) << endl;
  }
  return 0;
}

```