#include <bits/stdc++.h>
using namespace std;
int a, b, c, d, e, i, j, ii, jj, zx, xc, tes, t, pas, lef, rig, mid;
pair<int, int> p[200009], pr[200009];
deque<pair<int, int> > de;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> tes;
  for (t = 1; t <= tes; t++) {
    cin >> a;
    for (i = 1; i <= a; i++) {
      cin >> p[i].first >> p[i].second;
    }
    pas = 0;
    sort(p + 1, p + a + 1);
    while (de.size()) de.pop_back();
    pr[0].first = 0;
    pr[0].second = -1000000003;
    for (i = 1; i <= a; i++) {
      pr[i].first = pr[i - 1].first;
      if (pr[i - 1].second < p[i].first) {
        pr[i].first++;
      }
      pr[i].second = max(pr[i - 1].second, p[i].second);
    }
    for (i = a; i >= 1; i--) {
      zx = pr[i - 1].first + de.size();
      lef = -1;
      rig = de.size();
      c = -1;
      while (1) {
        if (lef + 1 >= rig) break;
        mid = (lef + rig) / 2;
        if (de[mid].first > pr[i - 1].second) {
          rig = mid;
        } else {
          c = mid;
          lef = mid;
        }
      }
      zx -= (c + 1);
      if (pas < zx) pas = zx;
      c = p[i].first;
      d = p[i].second;
      while (de.size()) {
        if (p[i].second < de[0].first) break;
        d = max(d, de[0].second);
        de.pop_front();
      }
      de.push_front(make_pair(c, d));
    }
    cout << pas << 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;
int a, b, c, d, e, i, j, ii, jj, zx, xc, tes, t, pas, lef, rig, mid;
pair<int, int> p[200009], pr[200009];
deque<pair<int, int> > de;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> tes;
  for (t = 1; t <= tes; t++) {
    cin >> a;
    for (i = 1; i <= a; i++) {
      cin >> p[i].first >> p[i].second;
    }
    pas = 0;
    sort(p + 1, p + a + 1);
    while (de.size()) de.pop_back();
    pr[0].first = 0;
    pr[0].second = -1000000003;
    for (i = 1; i <= a; i++) {
      pr[i].first = pr[i - 1].first;
      if (pr[i - 1].second < p[i].first) {
        pr[i].first++;
      }
      pr[i].second = max(pr[i - 1].second, p[i].second);
    }
    for (i = a; i >= 1; i--) {
      zx = pr[i - 1].first + de.size();
      lef = -1;
      rig = de.size();
      c = -1;
      while (1) {
        if (lef + 1 >= rig) break;
        mid = (lef + rig) / 2;
        if (de[mid].first > pr[i - 1].second) {
          rig = mid;
        } else {
          c = mid;
          lef = mid;
        }
      }
      zx -= (c + 1);
      if (pas < zx) pas = zx;
      c = p[i].first;
      d = p[i].second;
      while (de.size()) {
        if (p[i].second < de[0].first) break;
        d = max(d, de[0].second);
        de.pop_front();
      }
      de.push_front(make_pair(c, d));
    }
    cout << pas << endl;
  }
  return 0;
}

```