#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= 1000000007;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
long long dp[200001];
long long tot = 0;
void count(vector<pair<pair<long long, long long>, long long> > v) {
  long long t = 0;
  long long i = 0;
  tot = 0;
  set<long long> s;
  while (i < ((v).size())) {
    long long ele = v[i].first.first;
    if (t == 0) tot++;
    set<long long> temp, temp1;
    while (ele == v[i].first.first && i < ((v).size())) {
      if (v[i].second == 1)
        temp.insert(v[i].first.second);
      else
        temp1.insert(v[i].first.second);
      t += v[i].second;
      i++;
    }
    if (((s).size()) == 0 && ((temp).size()) == 1) dp[(*temp.begin())]--;
    for (auto x : temp) s.insert(x);
    for (auto x : temp1) s.erase(s.find(x));
  }
}
void fun(vector<pair<pair<long long, long long>, long long> > v) {
  long long i = 0;
  set<long long> s;
  while (i < ((v).size())) {
    long long ele = v[i].first.first;
    set<long long> temp, temp1;
    while (ele == v[i].first.first && i < ((v).size())) {
      if (v[i].second == 1)
        temp.insert(v[i].first.second);
      else
        temp1.insert(v[i].first.second);
      i++;
    }
    if (((s).size()) == 1 && ((temp).size()) != 0) dp[(*s.begin())]++;
    for (auto x : temp) s.insert(x);
    for (auto x : temp1) s.erase(s.find(x));
  }
}
void solve() {
  long long n;
  cin >> n;
  for (int i = 0; i < n; i++) dp[i] = 0;
  vector<pair<pair<long long, long long>, long long> > v;
  for (int i = 0; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    v.push_back(make_pair(make_pair(x, i), 1));
    v.push_back(make_pair(make_pair(y, i), -1));
  }
  sort(v.begin(), v.end());
  count(v);
  fun(v);
  long long ans = INT_MIN;
  for (int i = 0; i < n; i++) amax(ans, dp[i]);
  ans += tot;
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Your challenge is to write 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;
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= 1000000007;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
long long dp[200001];
long long tot = 0;
void count(vector<pair<pair<long long, long long>, long long> > v) {
  long long t = 0;
  long long i = 0;
  tot = 0;
  set<long long> s;
  while (i < ((v).size())) {
    long long ele = v[i].first.first;
    if (t == 0) tot++;
    set<long long> temp, temp1;
    while (ele == v[i].first.first && i < ((v).size())) {
      if (v[i].second == 1)
        temp.insert(v[i].first.second);
      else
        temp1.insert(v[i].first.second);
      t += v[i].second;
      i++;
    }
    if (((s).size()) == 0 && ((temp).size()) == 1) dp[(*temp.begin())]--;
    for (auto x : temp) s.insert(x);
    for (auto x : temp1) s.erase(s.find(x));
  }
}
void fun(vector<pair<pair<long long, long long>, long long> > v) {
  long long i = 0;
  set<long long> s;
  while (i < ((v).size())) {
    long long ele = v[i].first.first;
    set<long long> temp, temp1;
    while (ele == v[i].first.first && i < ((v).size())) {
      if (v[i].second == 1)
        temp.insert(v[i].first.second);
      else
        temp1.insert(v[i].first.second);
      i++;
    }
    if (((s).size()) == 1 && ((temp).size()) != 0) dp[(*s.begin())]++;
    for (auto x : temp) s.insert(x);
    for (auto x : temp1) s.erase(s.find(x));
  }
}
void solve() {
  long long n;
  cin >> n;
  for (int i = 0; i < n; i++) dp[i] = 0;
  vector<pair<pair<long long, long long>, long long> > v;
  for (int i = 0; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    v.push_back(make_pair(make_pair(x, i), 1));
    v.push_back(make_pair(make_pair(y, i), -1));
  }
  sort(v.begin(), v.end());
  count(v);
  fun(v);
  long long ans = INT_MIN;
  for (int i = 0; i < n; i++) amax(ans, dp[i]);
  ans += tot;
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```