#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 2e6 + 100;
const long long MOD = 998244353;
pair<int, int> pos[maxn];
int t, n, cover[maxn];
void solve() {
  int ans = 0;
  multiset<int> s;
  for (int i = 1; i <= 2 * n; i++) {
    if (pos[i].second > 0) {
      s.insert(pos[i].second);
    } else if (pos[i].second < 0) {
      s.erase(s.find(-pos[i].second));
    }
    if (s.size() == 1 && pos[i].second < 0 && pos[i + 1].second > 0 &&
        pos[i].first < pos[i + 1].first) {
      cover[*s.begin()]++;
    }
    if (s.size() == 1 && pos[i].second > 0 && pos[i + 1].second < 0) {
      cover[*s.begin()]--;
    }
    if (s.size() == 0) ans++;
  }
  int t = -INF;
  for (int i = 1; i <= n; i++) t = max(t, cover[i]);
  printf("%d\n", ans + t);
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    int tot = 0;
    for (int i = 1; i <= n; i++) {
      int l, r;
      scanf("%d", &l);
      scanf("%d", &r);
      pos[++tot] = make_pair(l, i);
      pos[++tot] = make_pair(r, -i);
      cover[i] = 0;
    }
    for (int i = 1; i <= tot; i++) pos[i].second *= -1;
    sort(pos + 1, pos + 1 + tot);
    for (int i = 1; i <= tot; i++) pos[i].second *= -1;
    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 INF = 0x3f3f3f3f;
const int maxn = 2e6 + 100;
const long long MOD = 998244353;
pair<int, int> pos[maxn];
int t, n, cover[maxn];
void solve() {
  int ans = 0;
  multiset<int> s;
  for (int i = 1; i <= 2 * n; i++) {
    if (pos[i].second > 0) {
      s.insert(pos[i].second);
    } else if (pos[i].second < 0) {
      s.erase(s.find(-pos[i].second));
    }
    if (s.size() == 1 && pos[i].second < 0 && pos[i + 1].second > 0 &&
        pos[i].first < pos[i + 1].first) {
      cover[*s.begin()]++;
    }
    if (s.size() == 1 && pos[i].second > 0 && pos[i + 1].second < 0) {
      cover[*s.begin()]--;
    }
    if (s.size() == 0) ans++;
  }
  int t = -INF;
  for (int i = 1; i <= n; i++) t = max(t, cover[i]);
  printf("%d\n", ans + t);
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    int tot = 0;
    for (int i = 1; i <= n; i++) {
      int l, r;
      scanf("%d", &l);
      scanf("%d", &r);
      pos[++tot] = make_pair(l, i);
      pos[++tot] = make_pair(r, -i);
      cover[i] = 0;
    }
    for (int i = 1; i <= tot; i++) pos[i].second *= -1;
    sort(pos + 1, pos + 1 + tot);
    for (int i = 1; i <= tot; i++) pos[i].second *= -1;
    solve();
  }
  return 0;
}

```