#include <bits/stdc++.h>
using namespace std;
const int N = 200006;
struct node {
  int l, r;
} A[N];
bool operator<(node x1, node x2) { return x1.l < x2.l; }
int n;
map<int, int> isl;
set<int> open;
struct node2 {
  int x, index, flag;
};
bool operator<(node2 x1, node2 x2) {
  if (x1.x == x2.x) {
    return x1.flag < x2.flag;
  }
  return x1.x < x2.x;
}
vector<node2> vec;
map<int, int> ans;
int ini = 0;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    isl.clear();
    open.clear();
    vec.clear();
    ans.clear();
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d%d", &A[i].l, &A[i].r);
    }
    sort(A + 1, A + n + 1);
    for (int i = 1; i <= n; i++) {
      vec.push_back(node2{A[i].l, i, -1});
      vec.push_back(node2{A[i].r, i, 1});
    }
    sort(vec.begin(), vec.end());
    int cl = A[1].l;
    isl[cl] = 1;
    int cr = A[1].r;
    for (int i = 2; i <= n; i++) {
      if (A[i].l > cr) {
        if (isl.find(A[i].l) == isl.end()) {
          isl[A[i].l] = 1;
        } else {
          isl[A[i].l]++;
        }
        cl = A[i].l;
        cr = A[i].r;
      } else {
        cr = max(cr, A[i].r);
      }
    }
    ini = isl.size();
    int len = vec.size();
    for (int i = 0; i < len; i++) {
      if (vec[i].flag == -1) {
        if (open.size() == 1) {
          set<int>::iterator iter = open.begin();
          if (ans.find(A[*iter].l) == ans.end()) {
            ans[A[*iter].l] = 1;
          } else {
            ans[A[*iter].l]++;
          }
        }
        open.insert(vec[i].index);
      } else {
        open.erase(vec[i].index);
      }
    }
    int mx = -2 * N;
    map<int, int>::iterator iter = ans.begin();
    for (; iter != ans.end(); iter++) {
      if (isl.find(iter->first) != isl.end() && isl[iter->first] == 1) {
        iter->second--;
      }
      mx = max(mx, iter->second);
    }
    if (ans.size() == 0) {
      printf("%d\n", ini - 1);
    } else {
      printf("%d\n", mx + ini);
    }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const int N = 200006;
struct node {
  int l, r;
} A[N];
bool operator<(node x1, node x2) { return x1.l < x2.l; }
int n;
map<int, int> isl;
set<int> open;
struct node2 {
  int x, index, flag;
};
bool operator<(node2 x1, node2 x2) {
  if (x1.x == x2.x) {
    return x1.flag < x2.flag;
  }
  return x1.x < x2.x;
}
vector<node2> vec;
map<int, int> ans;
int ini = 0;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    isl.clear();
    open.clear();
    vec.clear();
    ans.clear();
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d%d", &A[i].l, &A[i].r);
    }
    sort(A + 1, A + n + 1);
    for (int i = 1; i <= n; i++) {
      vec.push_back(node2{A[i].l, i, -1});
      vec.push_back(node2{A[i].r, i, 1});
    }
    sort(vec.begin(), vec.end());
    int cl = A[1].l;
    isl[cl] = 1;
    int cr = A[1].r;
    for (int i = 2; i <= n; i++) {
      if (A[i].l > cr) {
        if (isl.find(A[i].l) == isl.end()) {
          isl[A[i].l] = 1;
        } else {
          isl[A[i].l]++;
        }
        cl = A[i].l;
        cr = A[i].r;
      } else {
        cr = max(cr, A[i].r);
      }
    }
    ini = isl.size();
    int len = vec.size();
    for (int i = 0; i < len; i++) {
      if (vec[i].flag == -1) {
        if (open.size() == 1) {
          set<int>::iterator iter = open.begin();
          if (ans.find(A[*iter].l) == ans.end()) {
            ans[A[*iter].l] = 1;
          } else {
            ans[A[*iter].l]++;
          }
        }
        open.insert(vec[i].index);
      } else {
        open.erase(vec[i].index);
      }
    }
    int mx = -2 * N;
    map<int, int>::iterator iter = ans.begin();
    for (; iter != ans.end(); iter++) {
      if (isl.find(iter->first) != isl.end() && isl[iter->first] == 1) {
        iter->second--;
      }
      mx = max(mx, iter->second);
    }
    if (ans.size() == 0) {
      printf("%d\n", ini - 1);
    } else {
      printf("%d\n", mx + ini);
    }
  }
  return 0;
}

```