#include <bits/stdc++.h>
using namespace std;
vector<int> coordinate_compression(vector<long long> &a, int cur = 0) {
  int n = a.size();
  vector<long long> b(n);
  copy(a.begin(), a.end(), b.begin());
  sort(b.begin(), b.end());
  unordered_map<long long, int> mp;
  for (int i = 0; i < n; i++) {
    if ((i > 0) && b[i] == b[i - 1]) continue;
    mp[b[i]] = cur++;
  }
  vector<int> res(n);
  for (int i = 0; i < n; i++) {
    res[i] = mp[a[i]];
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    cin >> n;
    vector<pair<long long, long long>> a(n);
    for (int i = 0; i < n; i++) cin >> a[i].first >> a[i].second;
    vector<long long> b(2 * n);
    for (int i = 0; i < n; i++) {
      b[2 * i] = a[i].first;
      b[2 * i + 1] = a[i].second;
    }
    vector<int> bb = coordinate_compression(b);
    int m = 0;
    for (int i = 0; i < n; i++) {
      a[i].first = bb[2 * i] * 2;
      a[i].second = bb[2 * i + 1] * 2;
      m = max(m, (int)a[i].first);
      m = max(m, (int)a[i].second);
    }
    m += 5;
    vector<int> rui(m);
    for (int i = 0; i < n; i++) {
      rui[a[i].first]++;
      rui[a[i].second + 1]--;
    }
    for (int i = 1; i < m; i++) {
      rui[i] += rui[i - 1];
    }
    sort(a.begin(), a.end(),
         [&](pair<long long, long long> &p0, pair<long long, long long> &p1) {
           return p0.second < p1.second;
         });
    stack<long long> st;
    for (int i = 0; i < n; i++) {
      while (st.size() && st.top() >= a[i].first) st.pop();
      st.push(a[i].second);
    }
    int res = st.size();
    if (res == n) {
      cout << res - 1 << "\n";
      continue;
    }
    vector<int> co(m);
    int sum = 0;
    for (int i = 0; i < m; i++) {
      if (rui[i] == 1) sum++;
      co[i] = sum;
    }
    int mx = 0;
    for (int i = 0; i < n; i++) {
      int left = a[i].first + 1;
      int right = a[i].second - 1;
      if (rui[a[i].first] == 1) left++;
      if (rui[a[i].second] == 1) right--;
      if (left > right) continue;
      int te = co[right];
      if (left > 0) te -= co[left - 1];
      mx = max(mx, te);
    }
    res += mx;
    cout << res << "\n";
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
vector<int> coordinate_compression(vector<long long> &a, int cur = 0) {
  int n = a.size();
  vector<long long> b(n);
  copy(a.begin(), a.end(), b.begin());
  sort(b.begin(), b.end());
  unordered_map<long long, int> mp;
  for (int i = 0; i < n; i++) {
    if ((i > 0) && b[i] == b[i - 1]) continue;
    mp[b[i]] = cur++;
  }
  vector<int> res(n);
  for (int i = 0; i < n; i++) {
    res[i] = mp[a[i]];
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    cin >> n;
    vector<pair<long long, long long>> a(n);
    for (int i = 0; i < n; i++) cin >> a[i].first >> a[i].second;
    vector<long long> b(2 * n);
    for (int i = 0; i < n; i++) {
      b[2 * i] = a[i].first;
      b[2 * i + 1] = a[i].second;
    }
    vector<int> bb = coordinate_compression(b);
    int m = 0;
    for (int i = 0; i < n; i++) {
      a[i].first = bb[2 * i] * 2;
      a[i].second = bb[2 * i + 1] * 2;
      m = max(m, (int)a[i].first);
      m = max(m, (int)a[i].second);
    }
    m += 5;
    vector<int> rui(m);
    for (int i = 0; i < n; i++) {
      rui[a[i].first]++;
      rui[a[i].second + 1]--;
    }
    for (int i = 1; i < m; i++) {
      rui[i] += rui[i - 1];
    }
    sort(a.begin(), a.end(),
         [&](pair<long long, long long> &p0, pair<long long, long long> &p1) {
           return p0.second < p1.second;
         });
    stack<long long> st;
    for (int i = 0; i < n; i++) {
      while (st.size() && st.top() >= a[i].first) st.pop();
      st.push(a[i].second);
    }
    int res = st.size();
    if (res == n) {
      cout << res - 1 << "\n";
      continue;
    }
    vector<int> co(m);
    int sum = 0;
    for (int i = 0; i < m; i++) {
      if (rui[i] == 1) sum++;
      co[i] = sum;
    }
    int mx = 0;
    for (int i = 0; i < n; i++) {
      int left = a[i].first + 1;
      int right = a[i].second - 1;
      if (rui[a[i].first] == 1) left++;
      if (rui[a[i].second] == 1) right--;
      if (left > right) continue;
      int te = co[right];
      if (left > 0) te -= co[left - 1];
      mx = max(mx, te);
    }
    res += mx;
    cout << res << "\n";
  }
  return 0;
}

```