#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
template <class S, class T>
ostream& operator<<(ostream& os, const pair<S, T> v) {
  os << "(" << v.first << ", " << v.second << ")";
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T> v) {
  for (int i = 0; i < (int)v.size(); i++) {
    if (i > 0) {
      os << " ";
    }
    os << v[i];
  }
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<vector<T>> v) {
  for (int i = 0; i < (int)v.size(); i++) {
    if (i > 0) {
      os << endl;
    }
    os << v[i];
  }
  return os;
}
template <class T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T& a, const T& b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
string num2bit(long long num, long long len) {
  string bit = "";
  for (int i = 0; i < (int)len; ++i) {
    bit += char('0' + (num >> i & 1));
  }
  return bit;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> l(n), r(n), cl(n), cr(n);
  vector<long long> cand;
  for (int i = 0; i < (int)n; ++i) {
    cin >> l[i] >> r[i];
    r[i]++;
    cand.push_back(l[i]);
    cand.push_back(r[i]);
  }
  sort(cand.begin(), cand.end());
  cand.erase(unique(cand.begin(), cand.end()), cand.end());
  for (int i = 0; i < (int)n; ++i) {
    long long lpos = lower_bound(cand.begin(), cand.end(), l[i]) - cand.begin();
    cl[i] = 2 * lpos;
    long long rpos = lower_bound(cand.begin(), cand.end(), r[i]) - cand.begin();
    cr[i] = 2 * rpos - 1;
  }
  vector<long long> imos(5 * n + 1, 0), ip(5 * n + 1, 0), sip(5 * n + 2, 0);
  for (int i = 0; i < (int)n; ++i) {
    imos[cl[i]]++;
    imos[cr[i]]--;
  }
  for (int i = 0; i < (int)5 * n; ++i) imos[i + 1] += imos[i];
  long long somo = 0;
  for (int i = 0; i < (int)5 * n; ++i)
    if (imos[i] > 0 && imos[i + 1] == 0) somo++;
  for (int i = 0; i < (int)5 * n; ++i) {
    if (imos[i + 1] == 1 && imos[i] > 1) ip[i + 1] = 1;
  }
  for (int i = 0; i < (int)5 * n + 1; ++i) sip[i + 1] = sip[i] + ip[i];
  long long res = -1;
  for (int i = 0; i < (int)n; ++i) {
    long long left = cl[i];
    long long right = cr[i];
    long long val = sip[right] - sip[left];
    if (left > 0 && imos[left] == 1 && imos[left - 1] == 1) val++;
    if (imos[right] == 0 && imos[right - 1] == 1) val--;
    chmax(res, somo + val);
  }
  cout << res << endl;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Please formulate 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 long long mod = 1e9 + 7;
template <class S, class T>
ostream& operator<<(ostream& os, const pair<S, T> v) {
  os << "(" << v.first << ", " << v.second << ")";
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T> v) {
  for (int i = 0; i < (int)v.size(); i++) {
    if (i > 0) {
      os << " ";
    }
    os << v[i];
  }
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<vector<T>> v) {
  for (int i = 0; i < (int)v.size(); i++) {
    if (i > 0) {
      os << endl;
    }
    os << v[i];
  }
  return os;
}
template <class T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T& a, const T& b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
string num2bit(long long num, long long len) {
  string bit = "";
  for (int i = 0; i < (int)len; ++i) {
    bit += char('0' + (num >> i & 1));
  }
  return bit;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> l(n), r(n), cl(n), cr(n);
  vector<long long> cand;
  for (int i = 0; i < (int)n; ++i) {
    cin >> l[i] >> r[i];
    r[i]++;
    cand.push_back(l[i]);
    cand.push_back(r[i]);
  }
  sort(cand.begin(), cand.end());
  cand.erase(unique(cand.begin(), cand.end()), cand.end());
  for (int i = 0; i < (int)n; ++i) {
    long long lpos = lower_bound(cand.begin(), cand.end(), l[i]) - cand.begin();
    cl[i] = 2 * lpos;
    long long rpos = lower_bound(cand.begin(), cand.end(), r[i]) - cand.begin();
    cr[i] = 2 * rpos - 1;
  }
  vector<long long> imos(5 * n + 1, 0), ip(5 * n + 1, 0), sip(5 * n + 2, 0);
  for (int i = 0; i < (int)n; ++i) {
    imos[cl[i]]++;
    imos[cr[i]]--;
  }
  for (int i = 0; i < (int)5 * n; ++i) imos[i + 1] += imos[i];
  long long somo = 0;
  for (int i = 0; i < (int)5 * n; ++i)
    if (imos[i] > 0 && imos[i + 1] == 0) somo++;
  for (int i = 0; i < (int)5 * n; ++i) {
    if (imos[i + 1] == 1 && imos[i] > 1) ip[i + 1] = 1;
  }
  for (int i = 0; i < (int)5 * n + 1; ++i) sip[i + 1] = sip[i] + ip[i];
  long long res = -1;
  for (int i = 0; i < (int)n; ++i) {
    long long left = cl[i];
    long long right = cr[i];
    long long val = sip[right] - sip[left];
    if (left > 0 && imos[left] == 1 && imos[left - 1] == 1) val++;
    if (imos[right] == 0 && imos[right - 1] == 1) val--;
    chmax(res, somo + val);
  }
  cout << res << endl;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```