#include <bits/stdc++.h>
using namespace std;
bool isprime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i += 6) {
    if (n % i == 0 || n % (i + 2) == 0) return false;
  }
  return true;
}
long long modexp(long long a, long long b, long long m) {
  long long r = 1;
  a = a % m;
  while (b > 0) {
    if (b & 1) r = (r * a) % m;
    b = b >> 1;
    a = (a * a) % m;
  }
  return r % m;
}
long long builtin_popcount(long long x) {
  long long cnt = 0;
  for (long long i = 60; i >= 0; i--) {
    if ((1LL << i) & x) cnt++;
  }
  return cnt;
}
long long addmod(long long a, long long b) {
  a %= 998244353;
  b %= 998244353;
  return (a + b) % 998244353;
}
long long mulmod(long long a, long long b) {
  a %= 998244353;
  b %= 998244353;
  return (a * b) % 998244353;
}
long long submod(long long a, long long b) {
  a %= 998244353;
  b %= 998244353;
  a -= b;
  if (a < 0) a += 998244353;
  return a % 998244353;
}
struct CustomCompare {
  bool operator()(long long a, long long b) { return a > b; }
};
const int mxn = 2e9;
map<long long, long long> mp;
long long get_val(vector<pair<long long, long long>> arr) {
  long long ans = 0;
  long long l = -mxn, r = -mxn;
  sort((arr).begin(), (arr).end());
  for (int i = 0; i < arr.size(); i++) {
    if (arr[i].first > r) {
      if (l != -mxn) mp[l] = 0;
      l = arr[i].first, r = arr[i].second;
      ans++;
    } else
      r = max(r, arr[i].second);
  }
  mp[l] = 0;
  return ans;
}
void solve(vector<pair<long long, pair<long long, long long>>> query,
           vector<long long> &ans) {
  set<long long> open;
  for (int i = 0; i < query.size(); i++) {
    long long j = i - 1;
    vector<long long> op, cl;
    while (j + 1 < query.size() && query[i].first == query[j + 1].first) {
      j++;
      if (query[j].second.first == 1)
        op.push_back(query[j].second.second);
      else
        cl.push_back(query[j].second.second);
    }
    if (open.size() == 1 && !op.empty()) {
      ans[*open.begin()]++;
    }
    for (auto x : op) open.insert(x);
    for (auto x : cl) open.erase(x);
    i = j;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    mp.clear();
    long long n;
    cin >> n;
    vector<pair<long long, long long>> arr(n);
    for (int i = 0; i < n; i++) cin >> arr[i].first >> arr[i].second;
    vector<pair<long long, pair<long long, long long>>> query;
    for (int i = 0; i < n; i++) {
      query.push_back({arr[i].first, {1, i}});
      query.push_back({arr[i].second, {-1, i}});
    }
    sort((query).begin(), (query).end());
    vector<long long> ans(n);
    long long curr = get_val(arr);
    solve(query, ans);
    for (int i = 0; i < n; i++)
      if (mp.count(arr[i].first)) mp[arr[i].first]++;
    for (int i = 0; i < n; i++)
      if (mp[arr[i].first] == 1) ans[i]--;
    long long res = -mxn;
    for (auto x : ans) res = max(x, res);
    cout << curr + res << "\n";
  }
  return 0;
}

### 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;
bool isprime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i += 6) {
    if (n % i == 0 || n % (i + 2) == 0) return false;
  }
  return true;
}
long long modexp(long long a, long long b, long long m) {
  long long r = 1;
  a = a % m;
  while (b > 0) {
    if (b & 1) r = (r * a) % m;
    b = b >> 1;
    a = (a * a) % m;
  }
  return r % m;
}
long long builtin_popcount(long long x) {
  long long cnt = 0;
  for (long long i = 60; i >= 0; i--) {
    if ((1LL << i) & x) cnt++;
  }
  return cnt;
}
long long addmod(long long a, long long b) {
  a %= 998244353;
  b %= 998244353;
  return (a + b) % 998244353;
}
long long mulmod(long long a, long long b) {
  a %= 998244353;
  b %= 998244353;
  return (a * b) % 998244353;
}
long long submod(long long a, long long b) {
  a %= 998244353;
  b %= 998244353;
  a -= b;
  if (a < 0) a += 998244353;
  return a % 998244353;
}
struct CustomCompare {
  bool operator()(long long a, long long b) { return a > b; }
};
const int mxn = 2e9;
map<long long, long long> mp;
long long get_val(vector<pair<long long, long long>> arr) {
  long long ans = 0;
  long long l = -mxn, r = -mxn;
  sort((arr).begin(), (arr).end());
  for (int i = 0; i < arr.size(); i++) {
    if (arr[i].first > r) {
      if (l != -mxn) mp[l] = 0;
      l = arr[i].first, r = arr[i].second;
      ans++;
    } else
      r = max(r, arr[i].second);
  }
  mp[l] = 0;
  return ans;
}
void solve(vector<pair<long long, pair<long long, long long>>> query,
           vector<long long> &ans) {
  set<long long> open;
  for (int i = 0; i < query.size(); i++) {
    long long j = i - 1;
    vector<long long> op, cl;
    while (j + 1 < query.size() && query[i].first == query[j + 1].first) {
      j++;
      if (query[j].second.first == 1)
        op.push_back(query[j].second.second);
      else
        cl.push_back(query[j].second.second);
    }
    if (open.size() == 1 && !op.empty()) {
      ans[*open.begin()]++;
    }
    for (auto x : op) open.insert(x);
    for (auto x : cl) open.erase(x);
    i = j;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    mp.clear();
    long long n;
    cin >> n;
    vector<pair<long long, long long>> arr(n);
    for (int i = 0; i < n; i++) cin >> arr[i].first >> arr[i].second;
    vector<pair<long long, pair<long long, long long>>> query;
    for (int i = 0; i < n; i++) {
      query.push_back({arr[i].first, {1, i}});
      query.push_back({arr[i].second, {-1, i}});
    }
    sort((query).begin(), (query).end());
    vector<long long> ans(n);
    long long curr = get_val(arr);
    solve(query, ans);
    for (int i = 0; i < n; i++)
      if (mp.count(arr[i].first)) mp[arr[i].first]++;
    for (int i = 0; i < n; i++)
      if (mp[arr[i].first] == 1) ans[i]--;
    long long res = -mxn;
    for (auto x : ans) res = max(x, res);
    cout << curr + res << "\n";
  }
  return 0;
}

```