#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
const int N = 2e6 + 5;
int s[N];
vector<int> add[N], drop[N], own[N];
vector<pii> v;
int n, bnd;
static void solve() {
  map<int, int> mp;
  int ans, comp = 0, ptr = 0;
  int _cnt = 0, _xor = 0;
  cin >> n;
  v.resize(n);
  for (auto &i : v) {
    cin >> i.first >> i.second;
    i.first *= 2, i.second *= 2;
    mp[i.first] = mp[i.first + 1] = 0;
    mp[i.second] = mp[i.second + 1] = 0;
  }
  for (auto &it : mp) it.second = ptr++;
  bnd = mp.size() + 5;
  for (int i = 0; i < n; ++i) {
    v[i].first = mp[v[i].first];
    v[i].second = mp[v[i].second];
    add[v[i].first].push_back(i);
    drop[v[i].second].push_back(i);
  }
  for (auto i : v) {
    s[i.first] += 1;
    s[i.second + 1] -= 1;
  }
  for (int i = 1; i <= bnd; ++i) s[i] += s[i - 1];
  for (int i = 0; i <= bnd; ++i) {
    for (auto iv : add[i]) {
      _cnt += 1;
      _xor ^= iv;
    }
    if (_cnt == 1) own[_xor].push_back(i);
    for (auto iv : drop[i]) {
      _cnt -= 1;
      _xor ^= iv;
    }
  }
  for (int t = 0, i = 0; i <= bnd; ++i) {
    if (s[i] == 0) {
      if (t == 1) {
        comp += 1;
        t = 0;
      }
      continue;
    }
    t = 1;
  }
  ans = 0;
  for (int i = 0; i < n; ++i)
    if (!own[i].empty()) {
      vector<pii> range_stack;
      int shift = comp;
      for (auto j : own[i]) {
        if (range_stack.empty() || range_stack.back().second != j - 1)
          range_stack.emplace_back(j, j);
        else
          range_stack.back().second += 1;
      }
      for (auto j : range_stack) {
        if (j.first != 0 && s[j.first - 1] > 0 && s[j.second + 1] > 0)
          shift += 1;
      }
      if (range_stack.size() == 1 &&
          own[i].back() - own[i].front() == v[i].second - v[i].first)
        continue;
      ans = max(ans, shift);
    } else
      ans = max(ans, comp);
  for (int i = v[0].first; i <= v[0].second; ++i) s[i] -= 1;
  comp = 0;
  for (int t = 0, i = 0; i <= bnd; ++i) {
    if (s[i] == 0) {
      if (t == 1) {
        comp += 1;
        t = 0;
      }
      continue;
    }
    t = 1;
  }
  cout << max(ans, comp) << '\n';
}
static void clean() {
  for (int i = 0; i <= bnd + 5; ++i) {
    s[i] = 0;
    own[i].clear();
    add[i].clear();
    drop[i].clear();
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int tsk;
  cin >> tsk;
  while (tsk--) {
    solve();
    if (tsk > 0) clean();
  }
  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;
using pii = pair<int, int>;
const int N = 2e6 + 5;
int s[N];
vector<int> add[N], drop[N], own[N];
vector<pii> v;
int n, bnd;
static void solve() {
  map<int, int> mp;
  int ans, comp = 0, ptr = 0;
  int _cnt = 0, _xor = 0;
  cin >> n;
  v.resize(n);
  for (auto &i : v) {
    cin >> i.first >> i.second;
    i.first *= 2, i.second *= 2;
    mp[i.first] = mp[i.first + 1] = 0;
    mp[i.second] = mp[i.second + 1] = 0;
  }
  for (auto &it : mp) it.second = ptr++;
  bnd = mp.size() + 5;
  for (int i = 0; i < n; ++i) {
    v[i].first = mp[v[i].first];
    v[i].second = mp[v[i].second];
    add[v[i].first].push_back(i);
    drop[v[i].second].push_back(i);
  }
  for (auto i : v) {
    s[i.first] += 1;
    s[i.second + 1] -= 1;
  }
  for (int i = 1; i <= bnd; ++i) s[i] += s[i - 1];
  for (int i = 0; i <= bnd; ++i) {
    for (auto iv : add[i]) {
      _cnt += 1;
      _xor ^= iv;
    }
    if (_cnt == 1) own[_xor].push_back(i);
    for (auto iv : drop[i]) {
      _cnt -= 1;
      _xor ^= iv;
    }
  }
  for (int t = 0, i = 0; i <= bnd; ++i) {
    if (s[i] == 0) {
      if (t == 1) {
        comp += 1;
        t = 0;
      }
      continue;
    }
    t = 1;
  }
  ans = 0;
  for (int i = 0; i < n; ++i)
    if (!own[i].empty()) {
      vector<pii> range_stack;
      int shift = comp;
      for (auto j : own[i]) {
        if (range_stack.empty() || range_stack.back().second != j - 1)
          range_stack.emplace_back(j, j);
        else
          range_stack.back().second += 1;
      }
      for (auto j : range_stack) {
        if (j.first != 0 && s[j.first - 1] > 0 && s[j.second + 1] > 0)
          shift += 1;
      }
      if (range_stack.size() == 1 &&
          own[i].back() - own[i].front() == v[i].second - v[i].first)
        continue;
      ans = max(ans, shift);
    } else
      ans = max(ans, comp);
  for (int i = v[0].first; i <= v[0].second; ++i) s[i] -= 1;
  comp = 0;
  for (int t = 0, i = 0; i <= bnd; ++i) {
    if (s[i] == 0) {
      if (t == 1) {
        comp += 1;
        t = 0;
      }
      continue;
    }
    t = 1;
  }
  cout << max(ans, comp) << '\n';
}
static void clean() {
  for (int i = 0; i <= bnd + 5; ++i) {
    s[i] = 0;
    own[i].clear();
    add[i].clear();
    drop[i].clear();
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int tsk;
  cin >> tsk;
  while (tsk--) {
    solve();
    if (tsk > 0) clean();
  }
  return 0;
}

```