#include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int>> a(n);
  vector<vector<int>> k(2 * n, vector<int>(3));
  for (int i = 0; i < n; i++) {
    cin >> a[i].first >> a[i].second;
    k[i] = {a[i].first, 0, i};
    k[n + i] = {{a[i].second, 1, i}};
  }
  sort(k.begin(), k.end());
  vector<int> l(n), r(n);
  int ind = 0;
  for (auto x : k) {
    if (x[1] == 0)
      l[x[2]] = ind;
    else
      r[x[2]] = ind;
    ind++;
  }
  int cnt = 0;
  ind = 1;
  int mx = -1;
  int ans = 0;
  vector<int> prefix(3 * n + 1, 0);
  for (auto x : k) {
    if (x[1] == 0)
      cnt++;
    else
      cnt--;
    prefix[ind] = prefix[ind - 1] + ((cnt == 1) ? 1 : 0);
    if (x[1] == 1) {
      mx = max(prefix[r[x[2]]] - prefix[l[x[2]] + 1] + ((cnt == 0) ? -1 : 0),
               mx);
      ;
    }
    if (cnt == 0) ans++;
    ind++;
  }
  cout << ans + mx << "\n";
  ;
  ;
  ;
  ;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int>> a(n);
  vector<vector<int>> k(2 * n, vector<int>(3));
  for (int i = 0; i < n; i++) {
    cin >> a[i].first >> a[i].second;
    k[i] = {a[i].first, 0, i};
    k[n + i] = {{a[i].second, 1, i}};
  }
  sort(k.begin(), k.end());
  vector<int> l(n), r(n);
  int ind = 0;
  for (auto x : k) {
    if (x[1] == 0)
      l[x[2]] = ind;
    else
      r[x[2]] = ind;
    ind++;
  }
  int cnt = 0;
  ind = 1;
  int mx = -1;
  int ans = 0;
  vector<int> prefix(3 * n + 1, 0);
  for (auto x : k) {
    if (x[1] == 0)
      cnt++;
    else
      cnt--;
    prefix[ind] = prefix[ind - 1] + ((cnt == 1) ? 1 : 0);
    if (x[1] == 1) {
      mx = max(prefix[r[x[2]]] - prefix[l[x[2]] + 1] + ((cnt == 0) ? -1 : 0),
               mx);
      ;
    }
    if (cnt == 0) ans++;
    ind++;
  }
  cout << ans + mx << "\n";
  ;
  ;
  ;
  ;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```