#include <bits/stdc++.h>
using namespace std;
bool cmp(vector<int> v1, vector<int> v2) {
  if (v1[0] == v2[0]) {
    if (v1[2] == v2[2]) {
      if (v1[2] == -1)
        return v1[1] < v2[1];
      else
        return v2[1] < v1[1];
    }
    if (v1[2] == 1) {
      return true;
    } else {
      return false;
    }
  }
  return v1[0] < v2[0];
}
int main(int argc, const char* argv[]) {
  int n;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    vector<vector<int>> v;
    vector<vector<int>> vv;
    for (int j = 0; j < a; j++) {
      int b, c;
      cin >> b >> c;
      v.push_back({b, c, 1, j});
      v.push_back({c, b, -1, j});
      vv.push_back({b, c});
    }
    sort(v.begin(), v.end(), cmp);
    multiset<int> st;
    map<int, set<int>> m1;
    bool inter = false;
    bool end = false;
    for (int j = 0; j < v.size(); j++) {
      if (st.size() == 0) end = false;
      if (v[j][2] < 0) {
        st.erase(v[j][3]);
        if (st.size() == 1) end = true;
      } else {
        if (st.size() > 0) inter = true;
        if (st.size() == 1 && end) {
          if (m1.find(*st.begin()) == m1.end()) {
            m1[*st.begin()] = {};
          }
          m1[*st.begin()].insert(v[j][3]);
        }
        st.insert(v[j][3]);
      }
    }
    int mx = 0;
    int idx = -1;
    for (int j = 0; j < v.size(); j++) {
      if (m1[j].size() > mx) {
        mx = m1[j].size();
        idx = j;
      }
    }
    if (idx == -1) {
      if (!inter) vv.erase(vv.begin());
    } else {
      vv.erase(vv.begin() + idx);
    }
    sort(vv.begin(), vv.end());
    int cnt = 0;
    int r = -1000000009;
    for (int j = 0; j < vv.size(); j++) {
      if (vv[j][0] > r) {
        cnt++;
      }
      r = max(r, vv[j][1]);
    }
    cout << cnt << endl;
  }
  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 cmp(vector<int> v1, vector<int> v2) {
  if (v1[0] == v2[0]) {
    if (v1[2] == v2[2]) {
      if (v1[2] == -1)
        return v1[1] < v2[1];
      else
        return v2[1] < v1[1];
    }
    if (v1[2] == 1) {
      return true;
    } else {
      return false;
    }
  }
  return v1[0] < v2[0];
}
int main(int argc, const char* argv[]) {
  int n;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    vector<vector<int>> v;
    vector<vector<int>> vv;
    for (int j = 0; j < a; j++) {
      int b, c;
      cin >> b >> c;
      v.push_back({b, c, 1, j});
      v.push_back({c, b, -1, j});
      vv.push_back({b, c});
    }
    sort(v.begin(), v.end(), cmp);
    multiset<int> st;
    map<int, set<int>> m1;
    bool inter = false;
    bool end = false;
    for (int j = 0; j < v.size(); j++) {
      if (st.size() == 0) end = false;
      if (v[j][2] < 0) {
        st.erase(v[j][3]);
        if (st.size() == 1) end = true;
      } else {
        if (st.size() > 0) inter = true;
        if (st.size() == 1 && end) {
          if (m1.find(*st.begin()) == m1.end()) {
            m1[*st.begin()] = {};
          }
          m1[*st.begin()].insert(v[j][3]);
        }
        st.insert(v[j][3]);
      }
    }
    int mx = 0;
    int idx = -1;
    for (int j = 0; j < v.size(); j++) {
      if (m1[j].size() > mx) {
        mx = m1[j].size();
        idx = j;
      }
    }
    if (idx == -1) {
      if (!inter) vv.erase(vv.begin());
    } else {
      vv.erase(vv.begin() + idx);
    }
    sort(vv.begin(), vv.end());
    int cnt = 0;
    int r = -1000000009;
    for (int j = 0; j < vv.size(); j++) {
      if (vv[j][0] > r) {
        cnt++;
      }
      r = max(r, vv[j][1]);
    }
    cout << cnt << endl;
  }
  return 0;
}

```