#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
const int maxm = 1e6 + 10;
int n, ans[maxn];
pair<int, int> a[maxn];
void solve() {
  scanf("%d", &n);
  for (int l, r, i = 1; i <= n; ++i) {
    scanf("%d%d", &l, &r);
    a[i * 2 - 1] = {l, -i};
    a[i << 1] = {r, i};
    ans[i] = 0;
  }
  sort(a + 1, a + 1 + 2 * n);
  multiset<int> st;
  int group = 0;
  for (int i = 1; i <= 2 * n; ++i) {
    if (a[i].second < 0)
      st.insert(-a[i].second);
    else
      st.erase(st.find(a[i].second));
    if (st.empty())
      ++group;
    else if (st.size() == 1) {
      if (a[i].second > 0 && a[i + 1].second < 0 && a[i].first < a[i + 1].first)
        ++ans[*st.begin()];
      if (a[i].second < 0 && a[i + 1].second > 0) --ans[*st.begin()];
    }
  }
  int res = -n;
  for (int i = 1; i <= n; ++i) res = max(res, ans[i]);
  printf("%d\n", group + res);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}

### Prompt

Generate 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;
const int maxn = 1e6 + 10;
const int maxm = 1e6 + 10;
int n, ans[maxn];
pair<int, int> a[maxn];
void solve() {
  scanf("%d", &n);
  for (int l, r, i = 1; i <= n; ++i) {
    scanf("%d%d", &l, &r);
    a[i * 2 - 1] = {l, -i};
    a[i << 1] = {r, i};
    ans[i] = 0;
  }
  sort(a + 1, a + 1 + 2 * n);
  multiset<int> st;
  int group = 0;
  for (int i = 1; i <= 2 * n; ++i) {
    if (a[i].second < 0)
      st.insert(-a[i].second);
    else
      st.erase(st.find(a[i].second));
    if (st.empty())
      ++group;
    else if (st.size() == 1) {
      if (a[i].second > 0 && a[i + 1].second < 0 && a[i].first < a[i + 1].first)
        ++ans[*st.begin()];
      if (a[i].second < 0 && a[i + 1].second > 0) --ans[*st.begin()];
    }
  }
  int res = -n;
  for (int i = 1; i <= n; ++i) res = max(res, ans[i]);
  printf("%d\n", group + res);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}

```