#include <bits/stdc++.h>
using namespace std;
int t, l[800005], r[800005], b[800005], n, m, a[800005], ans;
map<int, int> mp;
set<int> sets;
set<int>::iterator it;
int main() {
  cin >> t;
  while (t--) {
    cin >> n;
    m = 0;
    mp.clear();
    sets.clear();
    memset(a, 0, sizeof(a));
    for (int i = 1; i <= n; i++) {
      cin >> l[i] >> r[i];
      b[++m] = l[i];
      b[++m] = r[i];
    }
    sort(b + 1, b + m + 1);
    m = unique(b + 1, b + m + 1) - b - 1;
    for (int i = 1; i <= m; i++) mp[b[i]] = i;
    for (int i = 1; i <= n; i++) {
      l[i] = mp[l[i]], r[i] = mp[r[i]];
      l[i] *= 2;
      l[i]--;
      r[i] *= 2;
      r[i]--;
      a[l[i]]++;
      a[r[i] + 1]--;
    }
    m *= 2;
    for (int i = 1; i <= m; i++) {
      a[i] = a[i - 1] + a[i];
      if (a[i] == 1) sets.insert(i);
    }
    int cnt = 0;
    for (int i = 1; i <= m; i++)
      if (a[i] && a[i + 1] == 0) cnt++;
    ans = cnt;
    if (ans == n) ans--;
    if (!sets.empty())
      for (int i = 1; i <= n; i++) {
        int mx = 0;
        it = sets.upper_bound(r[i]);
        it--;
        if (*it >= l[i] && *it <= r[i]) {
          for (int j = l[i]; j <= r[i] - 1; j++)
            if (a[j] == 1 && a[j + 1] != 1) mx++;
          if (a[l[i]] == 1) mx--;
          ans = max(cnt + mx, ans);
        }
      }
    cout << ans << endl;
  }
}

### Prompt

Please formulate 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 t, l[800005], r[800005], b[800005], n, m, a[800005], ans;
map<int, int> mp;
set<int> sets;
set<int>::iterator it;
int main() {
  cin >> t;
  while (t--) {
    cin >> n;
    m = 0;
    mp.clear();
    sets.clear();
    memset(a, 0, sizeof(a));
    for (int i = 1; i <= n; i++) {
      cin >> l[i] >> r[i];
      b[++m] = l[i];
      b[++m] = r[i];
    }
    sort(b + 1, b + m + 1);
    m = unique(b + 1, b + m + 1) - b - 1;
    for (int i = 1; i <= m; i++) mp[b[i]] = i;
    for (int i = 1; i <= n; i++) {
      l[i] = mp[l[i]], r[i] = mp[r[i]];
      l[i] *= 2;
      l[i]--;
      r[i] *= 2;
      r[i]--;
      a[l[i]]++;
      a[r[i] + 1]--;
    }
    m *= 2;
    for (int i = 1; i <= m; i++) {
      a[i] = a[i - 1] + a[i];
      if (a[i] == 1) sets.insert(i);
    }
    int cnt = 0;
    for (int i = 1; i <= m; i++)
      if (a[i] && a[i + 1] == 0) cnt++;
    ans = cnt;
    if (ans == n) ans--;
    if (!sets.empty())
      for (int i = 1; i <= n; i++) {
        int mx = 0;
        it = sets.upper_bound(r[i]);
        it--;
        if (*it >= l[i] && *it <= r[i]) {
          for (int j = l[i]; j <= r[i] - 1; j++)
            if (a[j] == 1 && a[j + 1] != 1) mx++;
          if (a[l[i]] == 1) mx--;
          ans = max(cnt + mx, ans);
        }
      }
    cout << ans << endl;
  }
}

```