#include <bits/stdc++.h>
using namespace std;
const int dx[8] = {0, -1, 0, 1, 1, -1, -1, 1},
          dy[8] = {-1, 0, 1, 0, -1, -1, 1, 1};
const long long mod = 1000000007;
const int base = 311;
const int N = 2e5 + 5;
vector<int> s;
int l[N], r[N];
int n;
int f[4 * N], sum[4 * N];
void gogo() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> l[i] >> r[i];
    s.push_back(l[i]);
    s.push_back(r[i]);
  }
  sort((s).begin(), (s).end());
  s.resize(unique((s).begin(), (s).end()) - s.begin());
  int mx = 0;
  for (int i = 1; i <= n; ++i) {
    l[i] = lower_bound((s).begin(), (s).end(), l[i]) - s.begin() + 1;
    r[i] = lower_bound((s).begin(), (s).end(), r[i]) - s.begin() + 1;
    l[i] *= 2;
    r[i] *= 2;
    mx = max(mx, r[i]);
    f[l[i]]++;
    f[r[i] + 1]--;
  }
  for (int i = 1; i <= mx; ++i) f[i] += f[i - 1];
  for (int i = 1; i <= mx; ++i) {
    sum[i] = sum[i - 1];
    if (f[i] == 1 && f[i - 1] >= 2) sum[i]++;
  }
  f[mx + 1] = 0;
  int cur = 0, ans = 0;
  for (int i = 1; i <= mx; ++i) cur += (f[i] >= 1 && f[i - 1] == 0);
  for (int i = 1; i <= n; ++i) {
    ans = max(ans, cur + sum[r[i]] - sum[l[i] - 1] -
                       (f[r[i]] == 1 && f[r[i] + 1] == 0));
  }
  cout << ans << '\n';
  for (int i = 1; i <= mx + 1; ++i) f[i] = sum[i] = 0;
  s.clear();
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("sol"
            ".inp",
            "r")) {
    freopen(
        "sol"
        ".inp",
        "r", stdin);
    freopen(
        "sol"
        ".out",
        "w", stdout);
  }
  int t;
  cin >> t;
  while (t--) gogo();
  return 0;
}

### Prompt

In cpp, your task is to solve 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 dx[8] = {0, -1, 0, 1, 1, -1, -1, 1},
          dy[8] = {-1, 0, 1, 0, -1, -1, 1, 1};
const long long mod = 1000000007;
const int base = 311;
const int N = 2e5 + 5;
vector<int> s;
int l[N], r[N];
int n;
int f[4 * N], sum[4 * N];
void gogo() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> l[i] >> r[i];
    s.push_back(l[i]);
    s.push_back(r[i]);
  }
  sort((s).begin(), (s).end());
  s.resize(unique((s).begin(), (s).end()) - s.begin());
  int mx = 0;
  for (int i = 1; i <= n; ++i) {
    l[i] = lower_bound((s).begin(), (s).end(), l[i]) - s.begin() + 1;
    r[i] = lower_bound((s).begin(), (s).end(), r[i]) - s.begin() + 1;
    l[i] *= 2;
    r[i] *= 2;
    mx = max(mx, r[i]);
    f[l[i]]++;
    f[r[i] + 1]--;
  }
  for (int i = 1; i <= mx; ++i) f[i] += f[i - 1];
  for (int i = 1; i <= mx; ++i) {
    sum[i] = sum[i - 1];
    if (f[i] == 1 && f[i - 1] >= 2) sum[i]++;
  }
  f[mx + 1] = 0;
  int cur = 0, ans = 0;
  for (int i = 1; i <= mx; ++i) cur += (f[i] >= 1 && f[i - 1] == 0);
  for (int i = 1; i <= n; ++i) {
    ans = max(ans, cur + sum[r[i]] - sum[l[i] - 1] -
                       (f[r[i]] == 1 && f[r[i] + 1] == 0));
  }
  cout << ans << '\n';
  for (int i = 1; i <= mx + 1; ++i) f[i] = sum[i] = 0;
  s.clear();
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("sol"
            ".inp",
            "r")) {
    freopen(
        "sol"
        ".inp",
        "r", stdin);
    freopen(
        "sol"
        ".out",
        "w", stdout);
  }
  int t;
  cin >> t;
  while (t--) gogo();
  return 0;
}

```