#include <bits/stdc++.h>
using namespace std;
bool contains(pair<int, int> a, pair<int, int> b) {
  return (a.first <= b.first && a.second >= b.second);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    vector<pair<int, int> > segs(n);
    vector<int> p(n), s(n);
    for (int i = 0; i < n; i++) {
      scanf("%d %d", &segs[i].first, &segs[i].second);
    }
    sort(segs.begin(), segs.end());
    p[0] = 1;
    int lol = segs[0].second;
    for (int i = 1; i < n; i++) {
      p[i] = p[i - 1];
      if (lol < segs[i].first) p[i]++;
      lol = max(lol, segs[i].second);
    }
    stack<int> S;
    for (int i = n - 1; i >= 0; i--) {
      while (!S.empty()) {
        if (contains(segs[i], segs[S.top()]))
          S.pop();
        else
          break;
      }
      if (S.empty())
        s[i] = 1;
      else {
        s[i] = 1 + s[S.top()];
        if (segs[i].second >= segs[S.top()].first) s[i]--;
      }
      S.push(i);
    }
    int ans = s[1];
    int maxR = segs[0].second;
    int reach = 1;
    for (int i = 1; i < n; i++) {
      reach = max(reach, i + 1);
      while (true) {
        if (reach == n) break;
        if (segs[reach].second <= maxR) {
          reach++;
        } else
          break;
      }
      int cur = p[i - 1];
      if (reach < n) {
        cur += s[reach];
        if (maxR >= segs[reach].first) cur--;
      }
      ans = max(ans, cur);
      maxR = max(maxR, segs[i].second);
    }
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
bool contains(pair<int, int> a, pair<int, int> b) {
  return (a.first <= b.first && a.second >= b.second);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    vector<pair<int, int> > segs(n);
    vector<int> p(n), s(n);
    for (int i = 0; i < n; i++) {
      scanf("%d %d", &segs[i].first, &segs[i].second);
    }
    sort(segs.begin(), segs.end());
    p[0] = 1;
    int lol = segs[0].second;
    for (int i = 1; i < n; i++) {
      p[i] = p[i - 1];
      if (lol < segs[i].first) p[i]++;
      lol = max(lol, segs[i].second);
    }
    stack<int> S;
    for (int i = n - 1; i >= 0; i--) {
      while (!S.empty()) {
        if (contains(segs[i], segs[S.top()]))
          S.pop();
        else
          break;
      }
      if (S.empty())
        s[i] = 1;
      else {
        s[i] = 1 + s[S.top()];
        if (segs[i].second >= segs[S.top()].first) s[i]--;
      }
      S.push(i);
    }
    int ans = s[1];
    int maxR = segs[0].second;
    int reach = 1;
    for (int i = 1; i < n; i++) {
      reach = max(reach, i + 1);
      while (true) {
        if (reach == n) break;
        if (segs[reach].second <= maxR) {
          reach++;
        } else
          break;
      }
      int cur = p[i - 1];
      if (reach < n) {
        cur += s[reach];
        if (maxR >= segs[reach].first) cur--;
      }
      ans = max(ans, cur);
      maxR = max(maxR, segs[i].second);
    }
    printf("%d\n", ans);
  }
  return 0;
}

```