#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e6 + 100;
int t;
int n;
vector<int> v;
int getid(int num) {
  return lower_bound(v.begin(), v.end(), num) - v.begin() + 1;
}
struct node {
  int a, b;
} save[maxn];
int in[maxn], out[maxn], sum[maxn], num[maxn];
int maxx = 0;
void init() {
  v.clear();
  for (int i = 1; i <= maxx; i++) {
    in[i] = out[i] = 0;
    sum[i] = 0;
    num[i] = 0;
  }
  maxx = 0;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    int x, y;
    for (int i = 1; i <= n; i++) {
      scanf("%d%d", &x, &y);
      x *= 2, y *= 2;
      v.push_back(x);
      v.push_back(y);
      v.push_back(x + 1);
      v.push_back(y + 1);
      v.push_back(x - 1);
      v.push_back(y - 1);
      save[i].a = x;
      save[i].b = y;
    }
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    for (int i = 1; i <= n; i++) {
      save[i].a = getid(save[i].a);
      save[i].b = getid(save[i].b);
    }
    for (int i = 1; i <= n; i++) {
      in[save[i].a]++;
      out[save[i].b]++;
    }
    maxx = v.size();
    int tmp = 0;
    for (int i = 1; i <= maxx; i++) {
      tmp += in[i];
      num[i] += tmp;
      tmp -= out[i];
    }
    for (int i = 1; i <= maxx; i++) {
      if (num[i] == 1 && num[i - 1] != 1)
        sum[i] = sum[i - 1] + 1;
      else
        sum[i] = sum[i - 1];
    }
    int base = 0;
    for (int i = 2; i <= maxx; i++)
      if (num[i - 1] == 0 && num[i]) base++;
    int res = base - 1;
    for (int i = 1; i <= n; i++) {
      x = save[i].a;
      y = save[i].b;
      int add = sum[y] - sum[x - 1];
      if (num[x - 1] == 0 && num[x] == 1) add--;
      if (num[y] == 1 && num[y + 1] == 0) add--;
      res = max(res, base + add);
    }
    printf("%d\n", res);
    init();
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
const int maxn = 2e6 + 100;
int t;
int n;
vector<int> v;
int getid(int num) {
  return lower_bound(v.begin(), v.end(), num) - v.begin() + 1;
}
struct node {
  int a, b;
} save[maxn];
int in[maxn], out[maxn], sum[maxn], num[maxn];
int maxx = 0;
void init() {
  v.clear();
  for (int i = 1; i <= maxx; i++) {
    in[i] = out[i] = 0;
    sum[i] = 0;
    num[i] = 0;
  }
  maxx = 0;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    int x, y;
    for (int i = 1; i <= n; i++) {
      scanf("%d%d", &x, &y);
      x *= 2, y *= 2;
      v.push_back(x);
      v.push_back(y);
      v.push_back(x + 1);
      v.push_back(y + 1);
      v.push_back(x - 1);
      v.push_back(y - 1);
      save[i].a = x;
      save[i].b = y;
    }
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    for (int i = 1; i <= n; i++) {
      save[i].a = getid(save[i].a);
      save[i].b = getid(save[i].b);
    }
    for (int i = 1; i <= n; i++) {
      in[save[i].a]++;
      out[save[i].b]++;
    }
    maxx = v.size();
    int tmp = 0;
    for (int i = 1; i <= maxx; i++) {
      tmp += in[i];
      num[i] += tmp;
      tmp -= out[i];
    }
    for (int i = 1; i <= maxx; i++) {
      if (num[i] == 1 && num[i - 1] != 1)
        sum[i] = sum[i - 1] + 1;
      else
        sum[i] = sum[i - 1];
    }
    int base = 0;
    for (int i = 2; i <= maxx; i++)
      if (num[i - 1] == 0 && num[i]) base++;
    int res = base - 1;
    for (int i = 1; i <= n; i++) {
      x = save[i].a;
      y = save[i].b;
      int add = sum[y] - sum[x - 1];
      if (num[x - 1] == 0 && num[x] == 1) add--;
      if (num[y] == 1 && num[y + 1] == 0) add--;
      res = max(res, base + add);
    }
    printf("%d\n", res);
    init();
  }
  return 0;
}

```