#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
const int oo = 1e9 + 4;
const int SZ = 2e5 + 4;
int q, n;
void read() { cin >> q; }
void solve() {
  while (q--) {
    cin >> n;
    vector<pii> o(n);
    vector<pair<int, pii> > h;
    int i = 0;
    for (auto &t : o) {
      cin >> t.first >> t.second;
      h.push_back({t.first, {0, i}});
      h.push_back({t.second, {1, i}});
      i++;
    }
    sort(h.begin(), h.end());
    set<int> in;
    vector<int> cnt(n, 0);
    vector<int> num(n, -oo);
    vector<bool> tic(n, 0);
    int last = 0, k0 = 0;
    i = 0;
    for (auto t : h) {
      if (!t.second.first) {
        if (in.size() == 1) {
          int v = *in.begin();
          if (last) cnt[v]++;
        }
        in.insert(t.second.second);
        num[t.second.second] = i;
      } else {
        in.erase(t.second.second);
        if (num[t.second.second] == i - 1) tic[t.second.second] = 1;
      }
      i++;
      last = t.second.first;
      k0 += in.empty();
    }
    int mx = -oo;
    for (int i = 0; i < n; i++) mx = max(mx, cnt[i] - (tic[i] && k0 > 1));
    cout << k0 + mx << endl;
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  read();
  solve();
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
using pii = pair<int, int>;
const int oo = 1e9 + 4;
const int SZ = 2e5 + 4;
int q, n;
void read() { cin >> q; }
void solve() {
  while (q--) {
    cin >> n;
    vector<pii> o(n);
    vector<pair<int, pii> > h;
    int i = 0;
    for (auto &t : o) {
      cin >> t.first >> t.second;
      h.push_back({t.first, {0, i}});
      h.push_back({t.second, {1, i}});
      i++;
    }
    sort(h.begin(), h.end());
    set<int> in;
    vector<int> cnt(n, 0);
    vector<int> num(n, -oo);
    vector<bool> tic(n, 0);
    int last = 0, k0 = 0;
    i = 0;
    for (auto t : h) {
      if (!t.second.first) {
        if (in.size() == 1) {
          int v = *in.begin();
          if (last) cnt[v]++;
        }
        in.insert(t.second.second);
        num[t.second.second] = i;
      } else {
        in.erase(t.second.second);
        if (num[t.second.second] == i - 1) tic[t.second.second] = 1;
      }
      i++;
      last = t.second.first;
      k0 += in.empty();
    }
    int mx = -oo;
    for (int i = 0; i < n; i++) mx = max(mx, cnt[i] - (tic[i] && k0 > 1));
    cout << k0 + mx << endl;
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  read();
  solve();
}

```