#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<pair<int, int> > seg(n);
    for (pair<int, int>& p : seg) {
      cin >> p.first >> p.second;
      p.first *= 2;
      p.second = 2 * p.second + 1;
    }
    map<int, int> compres;
    for (pair<int, int>& p : seg) {
      compres[p.first] = 0;
      compres[p.second] = 0;
    }
    int i = 0;
    for (auto it = compres.begin(); it != compres.end(); it++) {
      it->second = i++;
    }
    int m = compres.size();
    vector<int> left(m, 0), right(m, 0), at(m, 0);
    for (pair<int, int>& p : seg) {
      at[compres[p.first]]++;
      at[compres[p.second]]--;
    }
    int cur = 0;
    vector<int> cnt(m, 0);
    int cnt0 = 0;
    for (int i = 0; i < m; ++i) {
      if (cur == 1) {
        cnt[i] = cnt[i - 1] + 1;
      } else
        cnt[i] = cnt[i - 1];
      left[i] = cur;
      right[i] = cur + at[i];
      cur = right[i];
      if (cur == 0) cnt0++;
    }
    if (cur != 0) throw -1;
    int maxp = -1;
    for (pair<int, int>& pi : seg) {
      int p = -1;
      int l = compres[pi.first];
      int r = compres[pi.second];
      int ccnt = cnt[r] - cnt[l];
      if (r - l == 1) {
        if (ccnt == 1) {
          if (left[l] != 0 && right[r] != 0)
            p = 1;
          else if (left[l] == 0 && right[r] == 0)
            p = -1;
          else
            p = 0;
        } else {
          p = 0;
        }
      } else {
        p = ccnt;
        if (left[l] == 0 && right[l] == 1) p--;
        if (left[r] == 1 && right[r] == 0) p--;
      }
      maxp = max(maxp, p);
    }
    cout << cnt0 + maxp << "\n";
  }
}

### 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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<pair<int, int> > seg(n);
    for (pair<int, int>& p : seg) {
      cin >> p.first >> p.second;
      p.first *= 2;
      p.second = 2 * p.second + 1;
    }
    map<int, int> compres;
    for (pair<int, int>& p : seg) {
      compres[p.first] = 0;
      compres[p.second] = 0;
    }
    int i = 0;
    for (auto it = compres.begin(); it != compres.end(); it++) {
      it->second = i++;
    }
    int m = compres.size();
    vector<int> left(m, 0), right(m, 0), at(m, 0);
    for (pair<int, int>& p : seg) {
      at[compres[p.first]]++;
      at[compres[p.second]]--;
    }
    int cur = 0;
    vector<int> cnt(m, 0);
    int cnt0 = 0;
    for (int i = 0; i < m; ++i) {
      if (cur == 1) {
        cnt[i] = cnt[i - 1] + 1;
      } else
        cnt[i] = cnt[i - 1];
      left[i] = cur;
      right[i] = cur + at[i];
      cur = right[i];
      if (cur == 0) cnt0++;
    }
    if (cur != 0) throw -1;
    int maxp = -1;
    for (pair<int, int>& pi : seg) {
      int p = -1;
      int l = compres[pi.first];
      int r = compres[pi.second];
      int ccnt = cnt[r] - cnt[l];
      if (r - l == 1) {
        if (ccnt == 1) {
          if (left[l] != 0 && right[r] != 0)
            p = 1;
          else if (left[l] == 0 && right[r] == 0)
            p = -1;
          else
            p = 0;
        } else {
          p = 0;
        }
      } else {
        p = ccnt;
        if (left[l] == 0 && right[l] == 1) p--;
        if (left[r] == 1 && right[r] == 0) p--;
      }
      maxp = max(maxp, p);
    }
    cout << cnt0 + maxp << "\n";
  }
}

```