#include <bits/stdc++.h>
using namespace std;
template <typename elemType>
inline void Read(elemType &T) {
  elemType 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();
  T = (w ? -X : X);
}
int L[200010], R[200010], Sum[1600010], a[1600010];
int T, N;
const int INF = 1 << 30;
vector<int> HashData;
int ElemNum;
inline void Discretization() {
  HashData.push_back(-INF);
  sort(HashData.begin(), HashData.end());
  HashData.erase(unique(HashData.begin(), HashData.end()), HashData.end());
  ElemNum = HashData.size() - 1;
  return;
}
inline int GetPos(int Value) {
  return lower_bound(HashData.begin(), HashData.end(), Value) -
         HashData.begin();
}
int main() {
  Read(T);
  while (T--) {
    Read(N);
    HashData.clear();
    for (register int i = 1; i <= N; ++i) {
      Read(L[i]);
      Read(R[i]);
      HashData.push_back(L[i]);
      HashData.push_back(R[i]);
    }
    Discretization();
    fill(Sum, Sum + ElemNum * 4 + 1, 0);
    fill(a, a + ElemNum * 4 + 1, 0);
    for (register int i = 1; i <= N; ++i) {
      ++Sum[GetPos(L[i]) << 1];
      --Sum[(GetPos(R[i]) << 1) + 1];
    }
    for (register int i = 1; i <= ElemNum * 4; ++i) Sum[i] += Sum[i - 1];
    bool flag = false, flagB = false;
    int cnt = 0;
    for (register int i = 1; i <= ElemNum * 4; ++i) {
      if (!flagB && Sum[i]) {
        ++cnt;
        flagB = true;
      }
      if (flagB && !Sum[i]) flagB = false;
      if (!flag && Sum[i] == 1) {
        ++a[i];
        flag = true;
      }
      if (flag && Sum[i] != 1) flag = false;
    }
    for (register int i = 1; i <= ElemNum * 4; ++i) a[i] += a[i - 1];
    int Ans = -INF;
    for (register int i = 1; i <= N; ++i) {
      int l = GetPos(L[i]) << 1, r = GetPos(R[i]) << 1;
      int temp = a[r] - a[l - 1];
      if (Sum[l] == 1 && Sum[l - 1] != 1)
        --temp;
      else if (Sum[r] == 1)
        --temp;
      if (Sum[l] == 1 && Sum[r] == 1) --temp;
      Ans = max(Ans, temp);
    }
    printf("%d\n", Ans + cnt);
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
template <typename elemType>
inline void Read(elemType &T) {
  elemType 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();
  T = (w ? -X : X);
}
int L[200010], R[200010], Sum[1600010], a[1600010];
int T, N;
const int INF = 1 << 30;
vector<int> HashData;
int ElemNum;
inline void Discretization() {
  HashData.push_back(-INF);
  sort(HashData.begin(), HashData.end());
  HashData.erase(unique(HashData.begin(), HashData.end()), HashData.end());
  ElemNum = HashData.size() - 1;
  return;
}
inline int GetPos(int Value) {
  return lower_bound(HashData.begin(), HashData.end(), Value) -
         HashData.begin();
}
int main() {
  Read(T);
  while (T--) {
    Read(N);
    HashData.clear();
    for (register int i = 1; i <= N; ++i) {
      Read(L[i]);
      Read(R[i]);
      HashData.push_back(L[i]);
      HashData.push_back(R[i]);
    }
    Discretization();
    fill(Sum, Sum + ElemNum * 4 + 1, 0);
    fill(a, a + ElemNum * 4 + 1, 0);
    for (register int i = 1; i <= N; ++i) {
      ++Sum[GetPos(L[i]) << 1];
      --Sum[(GetPos(R[i]) << 1) + 1];
    }
    for (register int i = 1; i <= ElemNum * 4; ++i) Sum[i] += Sum[i - 1];
    bool flag = false, flagB = false;
    int cnt = 0;
    for (register int i = 1; i <= ElemNum * 4; ++i) {
      if (!flagB && Sum[i]) {
        ++cnt;
        flagB = true;
      }
      if (flagB && !Sum[i]) flagB = false;
      if (!flag && Sum[i] == 1) {
        ++a[i];
        flag = true;
      }
      if (flag && Sum[i] != 1) flag = false;
    }
    for (register int i = 1; i <= ElemNum * 4; ++i) a[i] += a[i - 1];
    int Ans = -INF;
    for (register int i = 1; i <= N; ++i) {
      int l = GetPos(L[i]) << 1, r = GetPos(R[i]) << 1;
      int temp = a[r] - a[l - 1];
      if (Sum[l] == 1 && Sum[l - 1] != 1)
        --temp;
      else if (Sum[r] == 1)
        --temp;
      if (Sum[l] == 1 && Sum[r] == 1) --temp;
      Ans = max(Ans, temp);
    }
    printf("%d\n", Ans + cnt);
  }
  return 0;
}

```