#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (a % b)
    return gcd(b, a % b);
  else
    return b;
}
long long int solve(vector<long long int> a, int b) {
  if (!(b + 1)) return -1;
  vector<long long int> nse;
  vector<long long int> se;
  long long int cur = (1LL << b);
  if (!a.size()) return -cur * 2;
  for (int i = 0; i < a.size(); i++)
    if (a[i] & cur)
      se.push_back(a[i]);
    else
      nse.push_back(a[i]);
  long long int ans1 = solve(nse, b - 1);
  long long int ans2 = solve(se, b - 1);
  return min(max(ans1, ans2 + cur), max(ans2, ans1 + cur));
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    vector<tuple<long long int, long long int, int>> seg;
    for (int i = 0; i < n; i++) {
      long long int a, b;
      cin >> a >> b;
      seg.push_back({a, b, i});
    }
    sort(seg.begin(), seg.end());
    set<int> s;
    priority_queue<pair<long long int, int>> pq;
    int rem[n], pre[n];
    memset(rem, 0, sizeof(rem));
    memset(pre, 0, sizeof(pre));
    int cnt = 0;
    for (int i = 0; i < n; i++) {
      while (!pq.empty() && -pq.top().first < get<0>(seg[i]))
        s.erase(pq.top().second), pq.pop();
      if (s.size() == 1) {
        rem[*(s.begin())]++;
      }
      if (!s.size()) cnt++, pre[get<2>(seg[i])] = 1;
      s.insert(get<2>(seg[i]));
      pq.push({-get<1>(seg[i]), get<2>(seg[i])});
    }
    if (cnt == n) {
      cout << n - 1 << "\n";
    } else {
      int ans = cnt;
      for (int i = 0; i < n; i++) ans = max(ans, cnt + rem[i] - pre[i]);
      cout << ans << "\n";
    }
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
long long int gcd(long long int a, long long int b) {
  if (a % b)
    return gcd(b, a % b);
  else
    return b;
}
long long int solve(vector<long long int> a, int b) {
  if (!(b + 1)) return -1;
  vector<long long int> nse;
  vector<long long int> se;
  long long int cur = (1LL << b);
  if (!a.size()) return -cur * 2;
  for (int i = 0; i < a.size(); i++)
    if (a[i] & cur)
      se.push_back(a[i]);
    else
      nse.push_back(a[i]);
  long long int ans1 = solve(nse, b - 1);
  long long int ans2 = solve(se, b - 1);
  return min(max(ans1, ans2 + cur), max(ans2, ans1 + cur));
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    vector<tuple<long long int, long long int, int>> seg;
    for (int i = 0; i < n; i++) {
      long long int a, b;
      cin >> a >> b;
      seg.push_back({a, b, i});
    }
    sort(seg.begin(), seg.end());
    set<int> s;
    priority_queue<pair<long long int, int>> pq;
    int rem[n], pre[n];
    memset(rem, 0, sizeof(rem));
    memset(pre, 0, sizeof(pre));
    int cnt = 0;
    for (int i = 0; i < n; i++) {
      while (!pq.empty() && -pq.top().first < get<0>(seg[i]))
        s.erase(pq.top().second), pq.pop();
      if (s.size() == 1) {
        rem[*(s.begin())]++;
      }
      if (!s.size()) cnt++, pre[get<2>(seg[i])] = 1;
      s.insert(get<2>(seg[i]));
      pq.push({-get<1>(seg[i]), get<2>(seg[i])});
    }
    if (cnt == n) {
      cout << n - 1 << "\n";
    } else {
      int ans = cnt;
      for (int i = 0; i < n; i++) ans = max(ans, cnt + rem[i] - pre[i]);
      cout << ans << "\n";
    }
  }
  return 0;
}

```