#include <bits/stdc++.h>
using namespace std;
const int INF = 2147483647;
const int INF2 = 0x3f3f3f3f;
const long long INF64 = 1e18;
const double INFD = 1e30;
const double EPS = 1e-6;
const double PI = 3.14159265;
const long long MOD = 1e9 + 7;
template <typename T>
inline T read() {
  T X = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
  return w ? -X : X;
}
const int MAXN = 200005;
int n, m, k;
int dr[4] = {1, 0, -1, 0};
int dc[4] = {0, 1, 0, -1};
int dr2[8] = {1, 1, 1, -1, -1, -1, 0, 0};
int dc2[8] = {1, 0, -1, 1, 0, -1, 1, -1};
int CASE = 1;
struct Point {
  int x, id;
  bool end;
  bool operator<(const Point& other) const {
    return x < other.x || (x == other.x && end < other.end);
  }
};
int cov[MAXN];
void solve() {
  vector<Point> invs;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    invs.push_back({l, i, false});
    invs.push_back({r, i, true});
    cov[i] = 0;
  }
  sort(invs.begin(), invs.end());
  set<int> covered;
  int seg = 0;
  for (const auto& v : invs) {
    if (!v.end) {
      if (covered.empty()) --cov[v.id];
      covered.insert(v.id);
    } else {
      covered.erase(v.id);
      if (covered.empty()) --cov[v.id];
    }
    if (covered.empty()) seg++;
    if (covered.size() == 1) ++cov[*covered.begin()];
  }
  int ans = -INF2;
  for (int i = 0; i < n; i++) ans = max(ans, cov[i]);
  printf("%d\n", seg + ans);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    solve();
  }
  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;
const int INF = 2147483647;
const int INF2 = 0x3f3f3f3f;
const long long INF64 = 1e18;
const double INFD = 1e30;
const double EPS = 1e-6;
const double PI = 3.14159265;
const long long MOD = 1e9 + 7;
template <typename T>
inline T read() {
  T X = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
  return w ? -X : X;
}
const int MAXN = 200005;
int n, m, k;
int dr[4] = {1, 0, -1, 0};
int dc[4] = {0, 1, 0, -1};
int dr2[8] = {1, 1, 1, -1, -1, -1, 0, 0};
int dc2[8] = {1, 0, -1, 1, 0, -1, 1, -1};
int CASE = 1;
struct Point {
  int x, id;
  bool end;
  bool operator<(const Point& other) const {
    return x < other.x || (x == other.x && end < other.end);
  }
};
int cov[MAXN];
void solve() {
  vector<Point> invs;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    invs.push_back({l, i, false});
    invs.push_back({r, i, true});
    cov[i] = 0;
  }
  sort(invs.begin(), invs.end());
  set<int> covered;
  int seg = 0;
  for (const auto& v : invs) {
    if (!v.end) {
      if (covered.empty()) --cov[v.id];
      covered.insert(v.id);
    } else {
      covered.erase(v.id);
      if (covered.empty()) --cov[v.id];
    }
    if (covered.empty()) seg++;
    if (covered.size() == 1) ++cov[*covered.begin()];
  }
  int ans = -INF2;
  for (int i = 0; i < n; i++) ans = max(ans, cov[i]);
  printf("%d\n", seg + ans);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    solve();
  }
  return 0;
}

```