#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    set<int> s;
    vector<pair<int, int>> sg(n);
    for (int i = 0; i < n; i++) {
      scanf("%d%d", &sg[i].first, &sg[i].second);
      s.insert(sg[i].first);
      s.insert(sg[i].second);
    }
    int now = 1;
    map<int, int> m;
    for (int a : s) m[a] = now++;
    vector<int> d(5 * n, 0);
    for (pair<int, int> &p : sg) {
      tie(p.first, p.second) = make_pair(2 * m[p.first], 2 * m[p.second]);
      d[p.first]++;
      d[p.second + 1]--;
    }
    vector<int> v(5 * n, 0);
    for (int i = 1; i < 5 * n; i++) v[i] = v[i - 1] + d[i];
    set<int> pos;
    for (int i = 0; i < 5 * n; i++)
      if (v[i] > 1) pos.insert(i);
    vector<int> dp(5 * n, 0);
    bool ok = 0;
    for (int i = 1; i < 5 * n - 1; i++) {
      dp[i] = dp[i - 1];
      if (v[i] == 1 && v[i + 1] > 1 && ok) dp[i]++;
      if (v[i] == 0) ok = 0;
      if (v[i] > 1) ok = 1;
    }
    auto solve = [&](int l, int r) -> int {
      if (l > r) return 0;
      if (v[l - 1] == 0 && v[r + 1] == 0)
        return -1;
      else if (v[l - 1] == 0 || v[r + 1] == 0)
        return 0;
      else
        return 1;
    };
    int res = 0;
    for (int i = 0; i < 5 * n - 1; i++) res += (v[i] == 0 && v[i + 1] > 0);
    int add = -n;
    for (pair<int, int> p : sg) {
      int l, r;
      tie(l, r) = p;
      if (pos.lower_bound(l) == pos.end() || (*pos.lower_bound(l) > r))
        add = max(add, solve(l, r));
      else {
        int cnt = 0;
        int a = (*pos.lower_bound(l));
        int b = (*prev(pos.upper_bound(r)));
        cnt += solve(l, a - 1);
        cnt += solve(b + 1, r);
        cnt += (dp[b] - dp[a]);
        add = max(add, cnt);
      }
    }
    printf("%d\n", res + add);
  }
  return 0;
}

### Prompt

Your task is to create 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;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    set<int> s;
    vector<pair<int, int>> sg(n);
    for (int i = 0; i < n; i++) {
      scanf("%d%d", &sg[i].first, &sg[i].second);
      s.insert(sg[i].first);
      s.insert(sg[i].second);
    }
    int now = 1;
    map<int, int> m;
    for (int a : s) m[a] = now++;
    vector<int> d(5 * n, 0);
    for (pair<int, int> &p : sg) {
      tie(p.first, p.second) = make_pair(2 * m[p.first], 2 * m[p.second]);
      d[p.first]++;
      d[p.second + 1]--;
    }
    vector<int> v(5 * n, 0);
    for (int i = 1; i < 5 * n; i++) v[i] = v[i - 1] + d[i];
    set<int> pos;
    for (int i = 0; i < 5 * n; i++)
      if (v[i] > 1) pos.insert(i);
    vector<int> dp(5 * n, 0);
    bool ok = 0;
    for (int i = 1; i < 5 * n - 1; i++) {
      dp[i] = dp[i - 1];
      if (v[i] == 1 && v[i + 1] > 1 && ok) dp[i]++;
      if (v[i] == 0) ok = 0;
      if (v[i] > 1) ok = 1;
    }
    auto solve = [&](int l, int r) -> int {
      if (l > r) return 0;
      if (v[l - 1] == 0 && v[r + 1] == 0)
        return -1;
      else if (v[l - 1] == 0 || v[r + 1] == 0)
        return 0;
      else
        return 1;
    };
    int res = 0;
    for (int i = 0; i < 5 * n - 1; i++) res += (v[i] == 0 && v[i + 1] > 0);
    int add = -n;
    for (pair<int, int> p : sg) {
      int l, r;
      tie(l, r) = p;
      if (pos.lower_bound(l) == pos.end() || (*pos.lower_bound(l) > r))
        add = max(add, solve(l, r));
      else {
        int cnt = 0;
        int a = (*pos.lower_bound(l));
        int b = (*prev(pos.upper_bound(r)));
        cnt += solve(l, a - 1);
        cnt += solve(b + 1, r);
        cnt += (dp[b] - dp[a]);
        add = max(add, cnt);
      }
    }
    printf("%d\n", res + add);
  }
  return 0;
}

```