#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int>> segs(n);
  vector<pair<int, pair<int, int>>> par1;
  vector<int> par;
  for (long long i = 0; i < n; i++) {
    cin >> segs[i].first >> segs[i].second;
    segs[i].second++;
    par1.push_back({segs[i].first, {1, i}});
    par1.push_back({segs[i].second, {0, i}});
  }
  sort(par1.begin(), par1.end());
  vector<pair<int, int>> toind(n);
  int prv = -1e9 - 5, cnt = 0, prvsgn = -1;
  for (auto ii : par1) {
    if (ii.first != prv || ii.second.first != prvsgn) {
      if (prv != -1e9 - 5) par.push_back(prvsgn == 1 ? cnt : -cnt);
      cnt = 0;
    }
    cnt++;
    prv = ii.first;
    prvsgn = ii.second.first;
    if (ii.second.first)
      toind[ii.second.second].first = par.size();
    else
      toind[ii.second.second].second = par.size();
  }
  if (prv != -1e9 - 5) par.push_back(prvsgn == 1 ? cnt : -cnt);
  vector<int> pre(2 * n, 0), pre1(2 * n, 0);
  cnt = 0;
  int m = par.size();
  for (long long i = 0; i < m; i++)
    pre[i] = (i ? pre[i - 1] : 0) + par[i], cnt += (pre[i] == 0);
  for (long long i = 0; i < m; i++)
    pre1[i] = (i ? pre1[i - 1] : 0) + (pre[i] == 1 ? 1 : 0);
  int bst = 0;
  for (long long i = 0; i < n; i++) {
    int l = toind[i].first + 1, r = toind[i].second - 1;
    if (l > r) bst = max(bst, cnt - 1);
    int tot = (pre1[r] - (l ? pre1[l - 1] : 0)) -
              (pre[r + 1] == 0 && par[r + 1] == -1);
    bst = max(bst, tot + cnt);
  }
  cout << bst << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Please create a solution in CPP 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;
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int>> segs(n);
  vector<pair<int, pair<int, int>>> par1;
  vector<int> par;
  for (long long i = 0; i < n; i++) {
    cin >> segs[i].first >> segs[i].second;
    segs[i].second++;
    par1.push_back({segs[i].first, {1, i}});
    par1.push_back({segs[i].second, {0, i}});
  }
  sort(par1.begin(), par1.end());
  vector<pair<int, int>> toind(n);
  int prv = -1e9 - 5, cnt = 0, prvsgn = -1;
  for (auto ii : par1) {
    if (ii.first != prv || ii.second.first != prvsgn) {
      if (prv != -1e9 - 5) par.push_back(prvsgn == 1 ? cnt : -cnt);
      cnt = 0;
    }
    cnt++;
    prv = ii.first;
    prvsgn = ii.second.first;
    if (ii.second.first)
      toind[ii.second.second].first = par.size();
    else
      toind[ii.second.second].second = par.size();
  }
  if (prv != -1e9 - 5) par.push_back(prvsgn == 1 ? cnt : -cnt);
  vector<int> pre(2 * n, 0), pre1(2 * n, 0);
  cnt = 0;
  int m = par.size();
  for (long long i = 0; i < m; i++)
    pre[i] = (i ? pre[i - 1] : 0) + par[i], cnt += (pre[i] == 0);
  for (long long i = 0; i < m; i++)
    pre1[i] = (i ? pre1[i - 1] : 0) + (pre[i] == 1 ? 1 : 0);
  int bst = 0;
  for (long long i = 0; i < n; i++) {
    int l = toind[i].first + 1, r = toind[i].second - 1;
    if (l > r) bst = max(bst, cnt - 1);
    int tot = (pre1[r] - (l ? pre1[l - 1] : 0)) -
              (pre[r + 1] == 0 && par[r + 1] == -1);
    bst = max(bst, tot + cnt);
  }
  cout << bst << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```