#include <bits/stdc++.h>
using namespace std;
void read(int &x) {
  int flag = 1;
  x = 0;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') flag = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  }
  x *= flag;
}
map<int, int> m1, m2;
int n, cnt, l[1000005], r[1000005], a[1000005], sum[1000005];
void prepare() {
  int i;
std:
  map<int, int>::iterator it;
  m1.clear();
  m2.clear();
  for (i = 1; i <= n; ++i) m1[l[i]] = m1[r[i]] = 1;
  for (it = m1.begin(); it != m1.end(); it++) m2[(*it).first] = ++cnt;
  for (i = 1; i <= n; ++i) {
    l[i] = (m2[l[i]] << 1) - 1;
    r[i] = (m2[r[i]] << 1) - 1;
  }
}
int main() {
  int t, num, ans, i;
  read(t);
  while (t--) {
    read(n);
    cnt = 0;
    for (i = 1; i <= n; ++i) {
      read(l[i]);
      read(r[i]);
    }
    prepare();
    cnt = (cnt << 1) - 1;
    for (i = 1; i <= n; ++i) {
      ++a[l[i]];
      --a[r[i] + 1];
    }
    for (i = 1; i <= cnt; ++i) a[i] += a[i - 1];
    num = 0;
    ans = -1000000000;
    for (i = 1; i <= cnt; ++i) {
      sum[i] = sum[i - 1];
      if (a[i] && !a[i - 1]) ++num;
      if (a[i] > 1 && a[i - 1] <= 1) ++sum[i];
    }
    for (i = 1; i <= n; ++i)
      ans = max(ans, sum[r[i]] - sum[l[i] - 1] +
                         ((a[l[i]] > 1) && (a[l[i] - 1] > 1)) - 1);
    printf("%d\n", num + ans);
    for (i = 0; i <= (cnt << 1); ++i) a[i] = sum[i] = 0;
  }
  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;
void read(int &x) {
  int flag = 1;
  x = 0;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') flag = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  }
  x *= flag;
}
map<int, int> m1, m2;
int n, cnt, l[1000005], r[1000005], a[1000005], sum[1000005];
void prepare() {
  int i;
std:
  map<int, int>::iterator it;
  m1.clear();
  m2.clear();
  for (i = 1; i <= n; ++i) m1[l[i]] = m1[r[i]] = 1;
  for (it = m1.begin(); it != m1.end(); it++) m2[(*it).first] = ++cnt;
  for (i = 1; i <= n; ++i) {
    l[i] = (m2[l[i]] << 1) - 1;
    r[i] = (m2[r[i]] << 1) - 1;
  }
}
int main() {
  int t, num, ans, i;
  read(t);
  while (t--) {
    read(n);
    cnt = 0;
    for (i = 1; i <= n; ++i) {
      read(l[i]);
      read(r[i]);
    }
    prepare();
    cnt = (cnt << 1) - 1;
    for (i = 1; i <= n; ++i) {
      ++a[l[i]];
      --a[r[i] + 1];
    }
    for (i = 1; i <= cnt; ++i) a[i] += a[i - 1];
    num = 0;
    ans = -1000000000;
    for (i = 1; i <= cnt; ++i) {
      sum[i] = sum[i - 1];
      if (a[i] && !a[i - 1]) ++num;
      if (a[i] > 1 && a[i - 1] <= 1) ++sum[i];
    }
    for (i = 1; i <= n; ++i)
      ans = max(ans, sum[r[i]] - sum[l[i] - 1] +
                         ((a[l[i]] > 1) && (a[l[i] - 1] > 1)) - 1);
    printf("%d\n", num + ans);
    for (i = 0; i <= (cnt << 1); ++i) a[i] = sum[i] = 0;
  }
  return 0;
}

```