#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <typename T>
using posteriority_queue = priority_queue<T, vector<T>, greater<T> >;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fLL;
const double EPS = 1e-8;
const int MOD = 1000000007;
const int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
template <typename T, typename U>
inline bool chmax(T &a, U b) {
  return a < b ? (a = b, true) : false;
}
template <typename T, typename U>
inline bool chmin(T &a, U b) {
  return a > b ? (a = b, true) : false;
}
template <typename T>
void unique(vector<T> &a) {
  a.erase(unique((a).begin(), (a).end()), a.end());
}
struct IOSetup {
  IOSetup() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  }
} iosetup;
void solve() {
  int n;
  cin >> n;
  vector<ll> l(n), r(n), comp(n * 2);
  for (int i = (0); i < (n); ++i) {
    cin >> l[i] >> r[i];
    l[i] *= 2;
    (r[i] *= 2) += 1;
    comp[i] = l[i];
    comp[n + i] = r[i];
  }
  comp.emplace_back(-LINF);
  comp.emplace_back(LINF);
  sort((comp).begin(), (comp).end());
  unique(comp);
  for (int i = (0); i < (n); ++i) {
    l[i] = lower_bound((comp).begin(), (comp).end(), l[i]) - comp.begin();
    r[i] = lower_bound((comp).begin(), (comp).end(), r[i]) - comp.begin();
  }
  int m = comp.size();
  vector<vector<int> > ins(m), del(m);
  for (int i = (0); i < (n); ++i) {
    ins[l[i]].emplace_back(i);
    del[r[i]].emplace_back(i);
  }
  vector<int> imos(m, 0);
  vector<vector<int> > pos(n);
  set<int> st;
  for (int i = (0); i < (m); ++i) {
    for (int e : del[i]) st.erase(e);
    for (int e : ins[i]) st.emplace(e);
    imos[i] = st.size();
    if (st.size() == 1) pos[*st.begin()].emplace_back(i);
  }
  int base = 0;
  for (int i = (1); i < (m); ++i) base += imos[i - 1] == 0 && imos[i] > 0;
  int ans = 0;
  for (int i = (0); i < (n); ++i) {
    int plus = 0, idx = 0;
    for (int j = 0; j < pos[i].size();) {
      int k = j + 1;
      while (k < pos[i].size() && pos[i][k] - pos[i][k - 1] == 1) ++k;
      plus += imos[pos[i][j] - 1] > 0 && imos[pos[i][k - 1] + 1] > 0;
      j = k;
    }
    chmax(ans, base + plus -
                   (imos[l[i] - 1] == 0 && imos[r[i]] == 0 &&
                    pos[i].size() == r[i] - l[i]));
  }
  cout << ans << '\n';
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Your task is to create 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;
using ll = long long;
template <typename T>
using posteriority_queue = priority_queue<T, vector<T>, greater<T> >;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fLL;
const double EPS = 1e-8;
const int MOD = 1000000007;
const int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
template <typename T, typename U>
inline bool chmax(T &a, U b) {
  return a < b ? (a = b, true) : false;
}
template <typename T, typename U>
inline bool chmin(T &a, U b) {
  return a > b ? (a = b, true) : false;
}
template <typename T>
void unique(vector<T> &a) {
  a.erase(unique((a).begin(), (a).end()), a.end());
}
struct IOSetup {
  IOSetup() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  }
} iosetup;
void solve() {
  int n;
  cin >> n;
  vector<ll> l(n), r(n), comp(n * 2);
  for (int i = (0); i < (n); ++i) {
    cin >> l[i] >> r[i];
    l[i] *= 2;
    (r[i] *= 2) += 1;
    comp[i] = l[i];
    comp[n + i] = r[i];
  }
  comp.emplace_back(-LINF);
  comp.emplace_back(LINF);
  sort((comp).begin(), (comp).end());
  unique(comp);
  for (int i = (0); i < (n); ++i) {
    l[i] = lower_bound((comp).begin(), (comp).end(), l[i]) - comp.begin();
    r[i] = lower_bound((comp).begin(), (comp).end(), r[i]) - comp.begin();
  }
  int m = comp.size();
  vector<vector<int> > ins(m), del(m);
  for (int i = (0); i < (n); ++i) {
    ins[l[i]].emplace_back(i);
    del[r[i]].emplace_back(i);
  }
  vector<int> imos(m, 0);
  vector<vector<int> > pos(n);
  set<int> st;
  for (int i = (0); i < (m); ++i) {
    for (int e : del[i]) st.erase(e);
    for (int e : ins[i]) st.emplace(e);
    imos[i] = st.size();
    if (st.size() == 1) pos[*st.begin()].emplace_back(i);
  }
  int base = 0;
  for (int i = (1); i < (m); ++i) base += imos[i - 1] == 0 && imos[i] > 0;
  int ans = 0;
  for (int i = (0); i < (n); ++i) {
    int plus = 0, idx = 0;
    for (int j = 0; j < pos[i].size();) {
      int k = j + 1;
      while (k < pos[i].size() && pos[i][k] - pos[i][k - 1] == 1) ++k;
      plus += imos[pos[i][j] - 1] > 0 && imos[pos[i][k - 1] + 1] > 0;
      j = k;
    }
    chmax(ans, base + plus -
                   (imos[l[i] - 1] == 0 && imos[r[i]] == 0 &&
                    pos[i].size() == r[i] - l[i]));
  }
  cout << ans << '\n';
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```