#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
const int inf = 1e9 + 7;
int test, n;
pair<int, int> a[maxn];
map<int, int> d;
void solve() {
  cin >> test;
  while (test--) {
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i].first >> a[i].second;
    sort(a + 1, a + n + 1);
    int l = -inf, r = -inf, ret = 0;
    for (int i = 1; i <= n; ++i)
      if (a[i].first > r) {
        if (l != -inf) d[l] = 0;
        ret++;
        l = a[i].first;
        r = a[i].second;
      } else
        r = max(r, a[i].second);
    d[l] = 0;
    vector<pair<int, pair<int, int> > > q;
    for (int i = 1; i <= n; ++i) {
      q.push_back(pair<int, pair<int, int> >(a[i].first, pair<int, int>(1, i)));
      q.push_back(
          pair<int, pair<int, int> >(a[i].second, pair<int, int>(-1, i)));
    }
    sort(q.begin(), q.end());
    set<int> op;
    vector<int> f(n + 1, 0);
    for (int i = 0, m = (int)q.size(); i < m; ++i) {
      int j = i;
      vector<int> be, en;
      while (j < m && q[j].first == q[i].first) {
        if (q[j].second.first == 1)
          be.push_back(q[j].second.second);
        else
          en.push_back(q[j].second.second);
        ++j;
      }
      if ((int)op.size() == 1 && !be.empty()) f[*op.begin()]++;
      for (auto &u : be) op.insert(u);
      for (auto &u : en) op.erase(u);
      i = j - 1;
    }
    for (int i = 1; i <= n; ++i)
      if (d.count(a[i].first)) d[a[i].first]++;
    for (int i = 1; i <= n; ++i)
      if (d[a[i].first] == 1) f[i]--;
    int opt = 0;
    for (int i = 1; i <= n; ++i) opt = max(opt, ret + f[i]);
    cout << opt << endl;
  }
}
int main() { solve(); }

### Prompt

Your challenge is to write a cpp solution to 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;
const int maxn = 2e5 + 5;
const int inf = 1e9 + 7;
int test, n;
pair<int, int> a[maxn];
map<int, int> d;
void solve() {
  cin >> test;
  while (test--) {
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i].first >> a[i].second;
    sort(a + 1, a + n + 1);
    int l = -inf, r = -inf, ret = 0;
    for (int i = 1; i <= n; ++i)
      if (a[i].first > r) {
        if (l != -inf) d[l] = 0;
        ret++;
        l = a[i].first;
        r = a[i].second;
      } else
        r = max(r, a[i].second);
    d[l] = 0;
    vector<pair<int, pair<int, int> > > q;
    for (int i = 1; i <= n; ++i) {
      q.push_back(pair<int, pair<int, int> >(a[i].first, pair<int, int>(1, i)));
      q.push_back(
          pair<int, pair<int, int> >(a[i].second, pair<int, int>(-1, i)));
    }
    sort(q.begin(), q.end());
    set<int> op;
    vector<int> f(n + 1, 0);
    for (int i = 0, m = (int)q.size(); i < m; ++i) {
      int j = i;
      vector<int> be, en;
      while (j < m && q[j].first == q[i].first) {
        if (q[j].second.first == 1)
          be.push_back(q[j].second.second);
        else
          en.push_back(q[j].second.second);
        ++j;
      }
      if ((int)op.size() == 1 && !be.empty()) f[*op.begin()]++;
      for (auto &u : be) op.insert(u);
      for (auto &u : en) op.erase(u);
      i = j - 1;
    }
    for (int i = 1; i <= n; ++i)
      if (d.count(a[i].first)) d[a[i].first]++;
    for (int i = 1; i <= n; ++i)
      if (d[a[i].first] == 1) f[i]--;
    int opt = 0;
    for (int i = 1; i <= n; ++i) opt = max(opt, ret + f[i]);
    cout << opt << endl;
  }
}
int main() { solve(); }

```